Рационал алгебралық өрнектерді тепе-тең түрлендіру (Рустюмова 1.2.1)

Шешуі: $${x^3} + {x^2}y + x{y^2} + {y^3} = \left( {{x^3} + {x^2}y} \right) + \left( {x{y^2} + {y^3}} \right) = $$ $$ = {x^2}(x + y) + {y^2}(x + y) = \left( {{x^2} + {y^2}} \right)(x + y)$$

Шешуі: $${x^3} – 2{x^2}y + {y^2}x – 2{y^3} = \left( {{x^3} + {y^2}x} \right) – \left( {2{x^2}y + 2{y^3}} \right) = $$ $$ = x\left( {{x^2} + {y^2}} \right) – 2y\left( {{x^2} + {y^2}} \right) = \left( {{x^2} + {y^2}} \right)(x – 2y)$$

Шешуі: $${x^4}{y^2} + 2{x^4} + {x^2}{y^2} + 2{x^2} = \left( {{x^4}{y^2} + {x^2}{y^2}} \right) + \left( {2{x^4} + 2{x^2}} \right) = $$ $$ = {x^2}{y^2}\left( {{x^2} + 1} \right) + 2{x^2}\left( {{x^2} + 1} \right) = \left( {{x^2}{y^2} + 2{x^2}} \right)\left( {{x^2} + 1} \right) = $$ $$ = {x^2}\left( {{y^2} + 2} \right)\left( {{x^2} + 1} \right)$$

Шешуі: $${2{x^2} + 7x – 4 = 2(x + 4)\left( {x – \dfrac{1}{2}} \right) = (x + 4)(2x – 1)}$$ $${2{x^2} + 7x – 4 = 0}$$ $${D = 49 + 32 = 81 = {9^2}}$$ $${x_1} = \dfrac{{ – 7 + 9}}{4} = \dfrac{1}{2};$$ $${x_2} = \dfrac{{ – 7 – 9}}{4} = \dfrac{{ – 16}}{4} = – 4.$$

Шешуі: $$4{x^2} + y – 2x – {y^2} = \left( {4{x^2} – {y^2}} \right) + (y – 2x) = $$ $$ = (2x – y)(2x + y) – (2x – y) = (2x – y)(2x + y – 1)$$

Шешуі: $${x^2} – 9x + 9y – {y^2} = \left( {{x^2} – {y^2}} \right) – (9x – 9y) = $$ $$ = (x – y)(x + y) – 9(x – y) = (x – y)(x + y – 9)$$

Шешуі: $$3ax – 2 – x + 6a = (3ax + 6a) – (2 + x) = $$ $$ = 3a(x + 2) – (x + 2) = (x + 2)(3a – 1)$$

Шешуі: $$4xy – 3 – 2y + 6x = (4xy – 2y) + (6x – 3) = $$ $$ = 2y(2x – 1) + 3(2x – 1) = (2x – 1)(2y + 3)$$

Шешуі: $$4{x^2} + 5x – 9{x^2} + 15x = \left( {4{x^2} – 9{x^2}} \right) + (5x + 15x) = $$ $$ = – 5{x^2} + 20x = 5x(4 – x)$$

Шешуі: $${2{x^2} – 7x + 5 = 2\left( {x – \dfrac{5}{2}} \right)(x – 1) = (2x – 5)(x – 1)}$$ $${D = 49 – 40 = 9 = {3^2}}$$ $${x_1} = \dfrac{{7 + 3}}{4} = \dfrac{{10}}{4} = \dfrac{5}{2};$$ $${x_2} = \dfrac{{7 – 3}}{4} = \dfrac{4}{4} = 1.$$

Шешуі: $${6{x^2} – 11x – 30 = 6\left( {x – \dfrac{{10}}{3}} \right)\left( {x + \dfrac{3}{2}} \right) = (3x – 10)(2x + 3)}$$ $${D = 121 + 720 = 841 = {{29}^2}}$$ $${x_1} = \dfrac{{11 + 29}}{{12}} = \dfrac{{40}}{{12}} = \dfrac{{10}}{3};$$ $${x_2} = \dfrac{{11 – 29}}{{12}} = – \dfrac{{18}}{{12}} = – \dfrac{3}{2}.$$

Шешуі: $$4{x^2} – 4{y^2} + 3x + 3y = \left( {4{x^2} – 4{y^2}} \right) + (3x + 3y) = $$ $$ = 4\left( {{x^2} – {y^2}} \right) + 3(x + y) = 4(x – y)(x + y) + 3(x + y) = $$ $$ = (x + y)(4(x – y) + 3) = (x + y)(4x – 4y + 3)$$

Шешуі: $${10{x^2} – 3x – 4 = 10\left( {x – \dfrac{4}{5}} \right)\left( {x + \dfrac{1}{2}} \right) = (5x – 4)(2x + 1)}$$ $${D = 9 + 160 = 169 = {{13}^2}}$$ $${x_1} = \frac{{3 + 13}}{{20}} = \frac{{16}}{{20}} = \frac{4}{5};$$ $${x_2} = \frac{{3 – 13}}{{20}} = – \frac{1}{2}.$$

Шешуі: $${2x + y + {y^2} – 4{x^2} = (2x + y) + \left( {{y^2} – 4{x^2}} \right) = }$$ $${ = (2x + y) + (y – 2x)(y + 2x) = (2x + y)(1 + y – 2x)}$$

Шешуі: $$5{x^2}y – 4x{y^2} – {y^3} = y\left( {5{x^2} – 4xy – {y^2}} \right) = $$ $$ = y\left( {4{x^2} – 4xy + {x^2} – {y^2}} \right) = y\left( {4x(x – y) + (x – y)(x + y)} \right) = $$ $$ = y(x – y)(4x + x + y) = y(x – y)(5x + y)$$

Шешуі: $${x^2}y – 4x{y^2} – 5{y^3} = y\left( {{x^2} – 4xy – 5{y^2}} \right) = $$ $$ = y\left( {{x^2} – {y^2} – 4xy – 4{y^2}} \right) = y\left( {(x – y)(x + y) – 4y(x + y)} \right) = $$ $$ = y(x + y)(x – y – 4y) = y(x + y)(x – 5y)$$

Шешуі: $${a^3} + 6{a^2} + 12a + 8 = {a^3} + 3{a^2} \cdot 2 + 3 \cdot a \cdot {2^2} + {2^3} = {(a + 2)^3}$$

Шешуі: $${4{a^2} – 12ab + 5{b^2} = 4{a^2} – 2ab – 10ab + 5{b^2} = }$$ $${ = 2a(2a – b) – 5b(2a – b) = (2a – b)(2a – 5b)}$$

Шешуі: $${{b^6} + {b^4}{c^2} – {b^2} – {c^2} = {b^4}\left( {{b^2} + {c^2}} \right) – \left( {{b^2} + {c^2}} \right) = }$$ $${ = \left( {{b^2} + {c^2}} \right)\left( {{b^4} – 1} \right) = \left( {{b^2} + {c^2}} \right)\left( {{b^2} + 1} \right)\left( {{b^2} – 1} \right) = \left( {{b^2} + {c^2}} \right)\left( {{b^2} + 1} \right)(b + 1)(b – 1)}$$

Шешуі: $$9{x^2} + 6xy + {y^2} – {z^2} = {(3x + y)^2} – {z^2} = $$ $$ = (3x + y – z)(3x + y + z)$$

Шешуі: $${(x + 1)^3} – 3{(x + 1)^2} + 3(x + 1) – 1 = {(x + 1 – 1)^3} = {x^3}$$

Шешуі: $$a – 3b + 9{b^2} – {a^2} = (a – 3b) + \left( {9{b^2} – {a^2}} \right) = $$ $$ = (a – 3b) – \left( {{a^2} – 9{b^2}} \right) = (a – 3b) – (a – 3b)(a + 3b) = $$ $$ = (a – 3b)(1 – a – 3b)$$

Шешуі: $${{x^2} + {y^2} + 4 + 2xy + 4x + 4y = \left( {{x^2} + 2xy + {y^2}} \right) + (4x + 4y) + 4 = }$$ $${ = {{(x + y)}^2} + 4(x + y) + 4 = {{(x + y + 2)}^2}}$$

Шешуі: $$2{a^4} + 30{a^3} + 150{a^2} + 250a = $$ $$ = 2a\left( {{a^3} + 15{a^2} + 75a + 125} \right) = 2a{(a + 5)^3}$$

Шешуі: $$64{(2 – 5a)^2} – 25{(6a – 5)^2} = $$ $$ = \left( {8(2 – 5a) + 5(6a – 5)} \right)\left( {8(2 – 5a) – 5(6a – 5)} \right) = $$ $$ = (16 – 40a + 30a – 25)(16 – 40a – 30a + 25) = $$ $$ = ( – 9 – 10a)(41 – 70a) = – (10a + 9) \cdot ( – (70a – 41)) = $$ $$ = (10a + 9)(70a – 41)$$