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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt `A=(1-3x)/(2x)+(3x-2)/(2x-1)+(3x-2)/(2x-4x^2)`
`=(2x(3x-2))/(2x(2x-1))-((3x-1)(2x-1))/(2x(2x-1))-(3x-2)/(2x(2x-1))`
`=(6x^2-4x-6x^2+5x-1-3x+2)/(2x(2x-1))`
`=(-2x+1)/(2x(2x-1))`
`=-1/(2x)`
`2x=1/(483)`
`=>A=-1/(1/483)=-483`
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\left(3x-2\right)^2+2\left(3x-2\right)\left(3x+2\right)+\left(3x+2\right)^2\)
\(=\left(3x-2+3x+2\right)^2\)
\(=36x^2\)(1)
Thay \(x=-\dfrac{1}{3}\) vào biểu thức (1), ta được:
\(36\cdot\left(-\dfrac{1}{3}\right)^2=36\cdot\dfrac{1}{9}=4\)
b) Sửa đề: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)
Ta có: \(\left(x+y-7\right)^2-2\cdot\left(x+y-7\right)\left(y-6\right)+\left(y-6\right)^2\)
\(=\left(x+y-7-y+6\right)^2\)
\(=\left(x-1\right)^2=100^2=10000\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
hay \(x=\dfrac{2}{7}\)
b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow x^3=-8\)
hay x=-2
Bài 1:
a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)
\(=xy^2-x^2y^2+x^2y^2-xy^2+xy\)
\(=xy\)
=1
b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)
\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)
\(=x^2-y^2\)
\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3-48x^2-36x-2x^3+18x+x^2-9\)
\(=-18x^3-46x^2-8x+16\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x-5\right)\left(2x+1\right)-2x\left(x-3\right)\\ =x.2x-5.2x+x-5-2x.x-2x.\left(-3\right)\\ =2x^2-10x+x-5-2x^2+6x\\ =2x^2-2x^2-10x+x+6x-5\\ =-3x-5\)
\(b,\left(2+3x\right)\left(2-3x\right)+\left(3x+4\right)^2\\ =\left[2^2-\left(3x\right)^2\right]+\left[\left(3x\right)^2+2.3x.4+4^2\right]\\=4-9x^2+\left(9x^2+24x+16\right)\\ =24x+20\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(3x\left(2x+1\right)+\left(2x-3\right)\left(x+1\right)\)
\(=6x^2+3x+2x^2+2x-3x-3\)
\(=8x^2+2x-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) A = (2x + 1)2 + (3x - 1)2 - 2(3x - 1)(2x + 1)
= (2x + 1 - 3x + 1)2 = (2 - x)2 = (x - 2)2
b) A = (x - 2)2 = (1002 - 2)2 = 10002 = 1000000
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{9x^2-6x+1}{9x^2-1}\)
\(=\dfrac{\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right)}\)
\(=\dfrac{3x-1}{3x+1}\)
\(=\dfrac{3\cdot\left(-3\right)-1}{3\cdot\left(-3\right)+1}=\dfrac{-9-1}{-9+1}=\dfrac{-10}{-8}=\dfrac{5}{4}\)
b) Ta có: \(\dfrac{x^2-6x+9}{3x^2-9x}\)
\(=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}\)
\(=\dfrac{x-3}{3x}\)
\(=\dfrac{-\dfrac{1}{3}-3}{3\cdot\dfrac{-1}{3}}=\dfrac{-\dfrac{10}{3}}{-1}=\dfrac{10}{3}\)
c) Ta có: \(\dfrac{x^2-4x+4}{2x^2-4x}\)
\(=\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}\)
\(=\dfrac{x-2}{2x}\)
\(=\dfrac{\dfrac{-1}{2}-2}{2\cdot\dfrac{-1}{2}}=\dfrac{-\dfrac{5}{2}}{-1}=\dfrac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa: \(A=\left(3x+2\right)^2+\left(2x-7\right)^2-2\left(3x+2\right)\left(2x-7\right)\)
\(A=\left(3x+2\right)^2-2\left(3x+2\right)\left(2x-7\right)+\left(2x-7\right)^2\)
\(A=\left[\left(3x+2\right)-\left(2x-7\right)\right]^2\)
\(A=\left(3x+2-2x+7\right)^2\)
\(A=\left(x+9\right)^2\)
Thay \(x=-19\) vào A ta có:
\(A=\left(-19+9\right)^2=\left(-10\right)^2=100\)
Vậy: ...
\(A=\left(3x+2\right)^2+\left(2-3x\right)^2-\left(1-2x\right)\left(2x+1\right)-\left(2x-3\right)\left(3x+2\right)\)
\(=\left(3x+2\right)^2-\left(2x-3\right)\left(3x+2\right)+\left(2-3x\right)^2+\left(2x-1\right)\left(2x+1\right)\)
\(=\left(3x+2-2+3x\right)^2+4x^2-1=\left(6x\right)^2+4x^2-1=40x^2-1\)
Ta có : \(\left|x\right|=\frac{1}{2}\Leftrightarrow x=\frac{1}{2};x=-\frac{1}{2}\)
Với x = 1/2 \(\frac{40.1}{4}-1=9\)
Với x = -1/2 \(\frac{40.1}{4}-1=9\)