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a) \(\left(-x-4\right)^2\)
\(=\left(-x\right)^2-2\cdot\left(-x\right)\cdot4+4^2\)
\(=x^2+8x+16\)
b) \(\left(-5+3x\right)^2\)
\(=\left(-5\right)^2+2\cdot\left(-5\right)\cdot3x+\left(3x\right)^2\)
\(=25-30x+9x^2\)
c) \(\left(-x-3\right)\left(x-3\right)\)
\(=-\left(x+3\right)\left(x-3\right)\)
\(=-\left(x^2-9\right)\)
m) \(\dfrac{1}{4}x^2-4x^2=\left(\dfrac{1}{2}x-2x\right)\left(\dfrac{1}{2}x+2x\right)\)
n) \(\dfrac{4}{49}-4x^2=\left(\dfrac{2}{7}-2x\right)\left(\dfrac{2}{7}+2x\right)\)
o) \(\left(x-3\right)\left(x+3\right)=x^2-9\)
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: \(N=\frac{2x-6}{x+3}\)
a ) \(\left(2x-3\right)^3=\left(2x\right)^3-3.\left(2x\right)^2.3+3.2x.3^2-3^3=8x^3-36x^2+54x-27\)
Có tổng hệ số là \(8-36+54-27=-1\)
b ) \(\left(x^2+2\right)^4=x^8+8x^6+24x^4+32x^2+16\)
Có tổng hệ số là : \(1+8+24+32+16=81\)
c ) \(\left(3x-5\right)^5=243x^5-2025x^4+6750x^3-11250x^2+9375x-3125\)
Có tổng hệ số là : \(243-2025+6750-11250+9375-3125=-32\)
a) \(\left(2x^3-y^2\right)^3\)
\(=\left(2x^3\right)^3-3\cdot\left(2x^3\right)^2\cdot y^2+3\cdot2x^3\cdot\left(y^2\right)^{^2}-\left(y^2\right)^3\)
\(=8x^9-3\cdot4x^6y^2+3\cdot2x^3y^4-y^6\)
\(=8x^9-12x^6y^2+6x^3y^4-y^6\)
b) \(\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
\(=x^3-\left(3y\right)^3\)
\(=x^3-27y^3\)
c) \(\left(x+2y+z\right)\left(x+2y-z\right)\)
\(=\left(x+2y\right)^2-z^2\)
\(=x^2+4xy+4y^2-z^2\)
d) \(\left(2x^3y-0,5x^2\right)^3\)
\(=\left(2x^3y-\dfrac{1}{2}x^2\right)^3\)
\(=8x^9y^3-6x^8y^2+\dfrac{3}{2}x^7y-\dfrac{1}{8}x^6\)
e) \(\left(x^2-3\right)\left(x^4+3x^2+9\right)\)
\(=\left(x^2-3\right)\left(4x^2+9\right)\)
\(=4x^4+9x^2-12x^2-27\)
\(=4x^4-3x^2-27\)
f) \(\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(=\left(2x\right)^3-1^3\)
\(=8x^3-1\)
\(a,\left(2x^3-y^2\right)^3=8x^9-12x^6y^2+6x^3y^4-y^6\)\(b,\left(x-3y\right)\left(x^2+3xy+9y^2\right)=x^3-27y^3\)
\(c,\left(x+2y+z\right)\left(x+2y-z\right)=\left(x+2y\right)^2-z^2=x^2+4xy+4y^2-z^2\)\(d,\left(2x^3y-0,5x^2\right)^3=8x^9y^3-6x^4y^2x^2+3x^3yx^4-0,125x^6=8x^9y^3-6x^6y^2+3x^7y-0,125x^6\)
Bài 3:
a) \(4x^2+4x+1=\left(2x+1\right)^2\)
b) \(9x^2-12x+4=\left(3x-2\right)^2\)
c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)
\(\left(x^3-1\right)\left(x^3+1\right)=\left(x^3\right)^2-1^2=x^6-1\)
\(\left(x^3-1\right)\left(x^3+1\right)\)
\(=\left(x^3\right)^2-1^2\)
\(=x^3-1\)
Z thôi T nha
a) Thiếu số mũ cuối cùng. Em bổ sung đề bài nhé.
b) \(B=\left(x-2\right)^3+\left(x+3\right)^4+\left(x-4\right)^3\)
\(=x^3-6x^2+12x-8+x^4+12x^3+54x^2+108x+81+x^3-12x^2+48x-64\\ =x^4+14x^3+36x^2+168x+9\)
Vậy hệ số của \(x^3\) là 14