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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: \(N=\frac{2x-6}{x+3}\)
\(x^2-6x+9=\left(x-3\right)^2\)
\(25+10x+x^2=\left(5+x\right)^2\)
\(\frac{1}{4}a^2+2ab^2+4b^4=\left(\frac{1}{2}a+2b^2\right)^2\)
\(\frac{1}{9}-\frac{2}{3}y^4+y^8=\left(\frac{1}{3}-y^4\right)^2\)
\(\left(3x+2\right)^2-4=\left(3x+2-2\right)\left(3x+2+2\right)=3x\left(3x+4\right)\)
\(4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)
\(4x^2-49=\left(2x-7\right)\left(2x+7\right)\)
\(8z^3+27=\left(2z+3\right)\left(4z^2-6z+9\right)\)
\(\frac{9}{25}x^4-\frac{1}{4}=\left(\frac{3}{5}x^2-\frac{1}{2}\right)\left(\frac{3}{5}x^2+\frac{1}{2}\right)\)
a) Rút gọn:
\(M=\left(x+3\right).\left(x^2-3x+9\right)-\left(x^3+54-x\right)\)
\(M=\left(x+3\right).\left(x^2-3x+3^2\right)-\left(x^3+54-x\right)\)
\(M=x^3+3^3-\left(x^3+54-x\right)\)
\(M=x^3+27-x^3-54+x\)
\(M=x-27.\)
+ Thay \(x=27\) vào biểu thức M ta được:
\(M=27-27\)
\(\Rightarrow M=0.\)
Vậy giá trị của biểu thức M tại \(x=27\) là: \(0.\)
Chúc bạn học tốt!
a) m m − 2 − m m + 2 m + 2 m m − 2 m = m + 2 m − 2
b) 3 5 − 3 m + 1 16 − m 2 m 2 + 2 m + 1 = 3 m − 12 5 ( m − 1 ) 16 − m 2 ( m + 1 ) 2 = − 3 ( m + 1 ) 5 ( m + 4 )
P = ( m2 - 2m + 4 )( m + 2 ) - m3( m + 3 )( m - 3 ) - m2 - 18
= m3 + 8 - m3( m2 - 9 ) - m2 - 18
= m3 + 8 - m5 + 9m3 - m2 - 18
= -m5 + 10m2 - m2 - 10
N = ( x + y )3 - 9( x + y )2 + 27( x + y ) - 27
= ( x + y )3 - 3.( x + y )2.3 + 3.( x + y ).32 - 33
= ( x + y - 3 )3
Phụ thuộc vào biến hết mà ;-;
\(P=\left(m^2-2m+4\right)\left(m+2\right)-m^3+\left(m+3\right)\left(m-3\right)-m^2-18\)
\(=m^3+8-m^3+\left(m^2-9\right)-m^2-18\)
\(=m^3+8-m^3+m^2-9-m^2-18\)
\(=-19\)
Vậy biểu thức trên kh thụ vào biến m
\(N=\left(x+y\right)^3-9\left(x+y\right)^2+27\left(x+y\right)-27\)
\(=\left(x+y\right)^3-3\left(x+y\right)^2.3+3\left(x+y\right)3^2-3^3\)
\(=\left(x+y-3\right)^3\)
c)\(\left(xy^2-1\right)\left(x^2y+5\right)\)
\(=x^3y^3+5xy^2-x^2y-5\)
d)\(4\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)\left(4x^2+1\right)\)
\(=4\left(x^2-\dfrac{1}{4}\right)\left(4x^2+1\right)\)
\(=4\left(4x^4+x^2-x-\dfrac{1}{4}\right)\)
\(=16x^4+4x^2-4x-1\)
( m − 2 ) ( m + 1 ) ( 2 m − 3 ) ( m + 1 ) = 2 m 2 − m − 6 ( 2 m − 3 ) ( 2 m + 3 ) = 4 m 3 − 2 m 2 − 3 m − 18 ( 2 m − 3 ) ( 4 m 2 + 6 m + 9 ) .