## Toán Lớp 8: Tìm x biết a)3x(x+1)-(x-2)(3x+1)=12 b)(3x ²-x+1)(x-1)+x ²(4-3x)=5/2 c)(x+3)(5x-1)=5(x+1)(x-2) d)(2x-3)(x+1)=2x(x-1) e)1/4x ²-(1/2x-4)1

Question

Toán Lớp 8: Tìm x biết
a)3x(x+1)-(x-2)(3x+1)=12
b)(3x ²-x+1)(x-1)+x ²(4-3x)=5/2
c)(x+3)(5x-1)=5(x+1)(x-2)
d)(2x-3)(x+1)=2x(x-1)
e)1/4x ²-(1/2x-4)1/2x=-14
f)3(1-4x)(x-1)+4(3x-2)(x+3)=-27
các bn làm hộ mk vs :)), hướng dẫn giải giúp em bài này ạ, em cảm ơn thầy cô và các bạn nhiều.

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1 giờ 2022-06-15T07:46:11+00:00 1 Answer 0 views 0

## TRẢ LỜI ( 1 )

1. Giải đáp:
$\begin{array}{l} a)3x\left( {x + 1} \right) – \left( {x – 2} \right)\left( {3x + 1} \right) = 12\\ \Leftrightarrow 3{x^2} + 3x – \left( {3{x^2} + x – 6x – 2} \right) = 12\\ \Leftrightarrow 3{x^2} + 3x – 3{x^2} + 5x + 2 = 12\\ \Leftrightarrow 8x = 10\\ \Leftrightarrow x = \dfrac{5}{4}\\ Vậy\,x = \dfrac{5}{4}\\ b)\left( {3{x^2} – x + 1} \right)\left( {x – 1} \right) + {x^2}\left( {4 – 3x} \right) = \dfrac{5}{2}\\ \Leftrightarrow 3{x^3} – 3{x^2} – {x^2} + x + x – 1\\ + 4{x^2} – 3{x^3} = \dfrac{5}{2}\\ \Leftrightarrow {x^2} + 2x + 1 = \dfrac{9}{2}\\ \Leftrightarrow {\left( {x + 1} \right)^2} = \dfrac{9}{2}\\ \Leftrightarrow x = \dfrac{{ – 2 \pm 3\sqrt 2 }}{2}\\ Vậy\,x = \dfrac{{ – 2 \pm 3\sqrt 2 }}{2}\\ c)\left( {x + 3} \right)\left( {5x – 1} \right) = 5\left( {x + 1} \right)\left( {x – 2} \right)\\ \Leftrightarrow 5{x^2} – x + 15x – 3 = 5\left( {{x^2} – x – 2} \right)\\ \Leftrightarrow 5{x^2} + 14x – 3 = 5{x^2} – 5x – 10\\ \Leftrightarrow 19x = – 7\\ \Leftrightarrow x = – \dfrac{7}{{19}}\\ Vậy\,x = – \dfrac{7}{{19}}\\ d)\left( {2x – 3} \right)\left( {x + 1} \right) = 2x\left( {x – 1} \right)\\ \Leftrightarrow 2{x^2} – x – 3 = 2{x^2} – 2x\\ \Leftrightarrow x = 3\\ Vậy\,x = 3\\ e)\dfrac{1}{4}{x^2} – \left( {\dfrac{1}{2}x – 4} \right).\dfrac{1}{2}x = – 14\\ \Leftrightarrow \dfrac{1}{4}{x^2} – \dfrac{1}{4}{x^2} + 2x = – 14\\ \Leftrightarrow x = – 7\\ Vậy\,x = – 7\\ f)3\left( {1 – 4x} \right)\left( {x – 1} \right) + 4\left( {3x – 2} \right)\left( {x + 3} \right) = – 27\\ \Leftrightarrow 3\left( {x – 1 – 4{x^2} + 4x} \right)\\ + 4\left( {3{x^2} + 9x – 2x – 6} \right) = – 27\\ \Leftrightarrow – 12{x^2} + 15x – 3\\ + 12{x^2} + 28x – 24 = – 27\\ \Leftrightarrow 43x = 0\\ \Leftrightarrow x = 0\\ Vậy\,x = 0 \end{array}$