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PH
1
16 tháng 7 2018
Bài 1:
\(\left(x+1\right)\left(x^2+2x-1\right)-x^2\left(x+3\right)=4\)
\(\Leftrightarrow x^3+2x^2-x+x^2+2x-1-x^3-3x^2=4\)
\(\Leftrightarrow x-1=4\)
\(\Leftrightarrow x=5\)
Bài 2:
\(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
\(=6x^2+9x+14x+21-6x^2-33x+10x+55\)
\(=76\)
Vậy bt A ko phụ thuộc vào biến x
\(B=\left(x^2-2\right)\left(x^2+x-1\right)-x\left(x^3+x^2-3x-2\right)\)
\(=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\)
\(=2\)
Vậy bt B ko phụ thuộc vào biến x
27 tháng 9 2018
\(a)\)\(VP=x^3+3x^2+2x\)
\(VP=x\left(x^2+3x+2\right)\)
\(VP=x\left[\left(x^2+x\right)+\left(2x+2\right)\right]\)
\(VP=x\left[x\left(x+1\right)+2\left(x+1\right)\right]\)
\(VP=x\left(x+1\right)\left(x+2\right)\) ( đpcm )
Chúc bạn học tốt ~
27 tháng 9 2018
a) x(x+1)(x+2)=(x2+x)(x+2)=x3+2x2+x2+2x=x3+3x2+3x
b)
(3x - 2)(4x - 5) - (2x - 1)(6x + 1) = 0
12x2 - 15x - 8x + 10 - 12x2 - 2x + 6x + 1 = 0
- 19x = - 11
x = 11/19
31 tháng 8 2020
a) ( x - 1 )3 + 3x( x - 1 )2 + 3x2( x - 1 ) + x3
= [ ( x - 1 ) + x ) ]3 ( HĐT số 4 )
= [ x - 1 + x ]3
= [ 2x - 1 ]3
=> đpcm
b) ( x2 - 2xy )3 + 3( x2 - 2xy )y2 + 3( x2 - 2xy )y4 + y6
= [ ( x2 - 2xy ) + y2 ]3 ( HĐT số 4 )
= [ x2 - 2xy + y2 ]3
= [ ( x - y )2 ]3
= ( x - y )6
=> đpcm
25 tháng 7 2018
\(a.\left(2x-3\right)\left(4x^2+6x+9\right)-\left(2x+3\right)\left(4x^2-6x+9\right)\\ =\left(2x\right)^3-3^3-\left[\left(2x\right)^3+3^3\right]\\ =8x^3-9-\left(8x^3+9\right)\\ =8x^3-9-8x^3-9=-18\)
\(b.\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\\ =x^3+1-\left(x^3-1\right)\\ =x^3+1-x^3+1=2\)
\(c.\left(3x-1\right)\left(3x+1\right)-\left(3x-2\right)^2\\ =9x^2-1-\left(9x^2-12x+4\right)\\ =9x^2-1-9x^2+12x-4\\ =12x-5\)
\(d.\left(2x-3\right)^2-\left(2x+3\right)\left(2x-3\right)\\ =\left(2x-3\right)\cdot\left[\left(2x-3\right)-\left(2x+3\right)\right]\\ =\left(2x-3\right)\cdot\left(2x-3-2x-3\right)\\ =\left(2x-3\right)\cdot\left(-6\right)\\ =-12x\cdot18\)
\(e.\left(3x-4\right)^2-\left(2x+4\right)^2\\ =9x^2-24x+16-\left(4x^2+16x+16\right)\\ =9x^2-24x+16-4x^2-16x-16\\ =5x^2-40x\)
\(f.\left(3x-5\right)^3-\left(3x+5\right)^3\\ =27x^3-135x^2+225x-125-\left(27x^3+135x^2+225x+125\right)\\ =27x^3-135x^2+225x-125-27x^3-135x^2-225x-125\\ =-270x^2-250\)
\(g.\left(2x-1\right)^2-\left(3x-1\right)^2\\ =4x^2-4x+1-\left(9x^2-6x+1\right)\\ =4x^2-4x+1-9x^2+6x-1\\ =-5x^2+2x\)
\(h.\left(x-2y\right)\left(x^2+2xy+4y^2\right)+\left(x^3-6y^3\right)\\ =x^3-8y^3+x^3-6y^3\\ =2x^3-14y^3\)
29 tháng 8 2020
Bài 1:
a) Ta có: \(VT=\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(u^2-3u+2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(n^2-u-2u+2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left[u\left(u-1\right)-2\left(u-1\right)\right]}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{-\left(u-1\right)\left(u-2\right)}{\left(u+2\right)\left(u-1\right)}\)
\(=\frac{2-u}{u+2}\)(1)
Ta có: \(VP=\frac{u^2-4u+4}{4-u^2}\)
\(=\frac{\left(u-2\right)^2}{-\left(u-2\right)\left(u+2\right)}\)
\(=\frac{-\left(u-2\right)}{u+2}\)
\(=\frac{2-u}{u+2}\)(2)
Từ (1) và (2) suy ra \(\frac{-u^2+3u-2}{\left(u+2\right)\left(u-1\right)}=\frac{u^2-4u+4}{4-u^2}\)
b) Ta có: \(VT=\frac{v^3+27}{v^2-3v+9}\)
\(=\frac{\left(v+3\right)\left(v^3-3u+9\right)}{v^2-3u+9}\)
\(=v+3=VP\)(đpcm)
Bài 2:
a) Ta có: \(\frac{3x^2-2x-5}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{3x^2-5x+3x-5}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{x\left(3x-5\right)+\left(3x-5\right)}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow\frac{\left(3x-5\right)\left(x+1\right)}{M}=\frac{3x-5}{2x-3}\)
\(\Leftrightarrow M=\frac{\left(3x-5\right)\left(x+1\right)\left(2x-3\right)}{3x-5}\)
\(\Leftrightarrow M=\left(x+1\right)\left(2x-3\right)\)
\(\Leftrightarrow M=2x^2-3x+2x-3\)
hay \(M=2x^2-x-3\)
Vậy: \(M=2x^2-x-3\)
b) Ta có: \(\frac{2x^2+3x-2}{x^2-4}=\frac{M}{x^2-4x+4}\)
\(\Leftrightarrow\frac{2x^2+4x-x-2}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{2x\left(x+2\right)-\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(2x-1\right)}{\left(x+2\right)\left(x-2\right)}=\frac{M}{\left(x-2\right)^2}\)
\(\Leftrightarrow\frac{M}{\left(x-2\right)^2}=\frac{2x-1}{x-2}\)
\(\Leftrightarrow M=\frac{\left(2x-1\right)\left(x-2\right)^2}{\left(x-2\right)}\)
\(\Leftrightarrow M=\left(2x-1\right)\left(x-2\right)\)
\(\Leftrightarrow M=2x^2-4x-x+2\)
hay \(M=2x^2-5x+2\)
Vậy: \(M=2x^2-5x+2\)
Bài 3:
a) Ta có: \(\frac{x+1}{N}=\frac{x^2-2x+4}{x^3+8}\)
\(\Leftrightarrow\frac{x+1}{N}=\frac{x^2-2x+4}{\left(x+2\right)\left(x^2-2x+4\right)}\)
\(\Leftrightarrow\frac{x+1}{N}=\frac{1}{x+2}\)
\(\Leftrightarrow N=\left(x+1\right)\left(x+2\right)\)
hay \(N=x^2+3x+2\)
Vậy: \(N=x^2+3x+2\)
n) Ta có: \(\frac{\left(x-3\right)\cdot N}{3+x}=\frac{2x^3-8x^2-6x+36}{2+x}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{2x^3+4x^2-12x^2-24x+18x+36}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{\left(x+3\right)}=\frac{2x^2\left(x+2\right)-12x\left(x+2\right)+18\left(x+2\right)}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=\frac{\left(x+2\right)\left(2x^2-12x+18\right)}{x+2}\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-12x+18\)
\(\Leftrightarrow\frac{N\cdot\left(x-3\right)}{x+3}=2x^2-6x-6x+18=2x\left(x-3\right)-6\left(x-3\right)=2\cdot\left(x-3\right)^2\)
\(\Leftrightarrow N\cdot\left(x-3\right)=\frac{2\left(x-3\right)^2}{x+3}\)
\(\Leftrightarrow N=\frac{2\left(x-3\right)^2}{x+3}:\left(x-3\right)=\frac{2\left(x-3\right)^2}{\left(x+3\right)\left(x-3\right)}\)
\(\Leftrightarrow N=\frac{2\left(x-3\right)}{x+3}\)
hay \(N=\frac{2x-6}{x+3}\)
Vậy: \(N=\frac{2x-6}{x+3}\)
ta có :