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Bài 1 :
a ) \(2x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(2x+2\right)=2\left(x+1\right)^2\)
b ) \(y^2\left(x^2+y\right)-zx^2-zy=y^2\left(x^2+y\right)-z\left(x^2+y\right)=\left(x^2+y\right)\left(y^2-z\right)\)
c ) \(4x\left(x-2y\right)+8y\left(2y-x\right)=4x\left(x-2y\right)-8y\left(x-2y\right)=4\left(x-2y\right)^2\)
d ) \(3x\left(x+1\right)^2-5x^2\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(3x^2+3x-5x^2+7\right)=\left(x+1\right)\left(3x-2x^2+7\right)\)
e ) \(x^2-6xy+9y^2=\left(x-3x\right)^2\)
Bài 1 :
f ) \(x^3+6x^2y+12xy^2+8y^3=\left(x+2y\right)^3\)
g ) \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
h ) \(125x^3+y^6=\left(5x+y^2\right)\left(25x^2-5xy^2+y^4\right)\)
\(1.\)
\(a,\left(a+b\right)^2=a^2+2ab+b^2\)
\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+2ab+b^2\)
\(\Rightarrow\left(a+b\right)^2=\left(a-b\right)^2+4ab\left(đpcm\right)\)
a) \(x^2+x+1=x^2+x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)(luôn dương)
b) \(x^2-x+\frac{1}{2}=x^2-x+\frac{1}{4}+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\frac{1}{4}>0\)(luôn dương)
Bài 1:
a) Ta có: \(x^2y^2-1\)
\(=\left(xy\right)^2-1^2\)
\(=\left(xy-1\right)\left(xy+1\right)\)
b) Ta có: \(x^4y^4-z^4\)
\(=\left(x^2y^2\right)^2-\left(z^2\right)^2\)
\(=\left(x^2y^2-z^2\right)\left(x^2y^2+z^2\right)\)
\(=\left(xy-z\right)\left(x^2y^2+z^2\right)\left(xy+z\right)\)
c) Ta có: \(\left(x+a\right)^2-25\)
\(=\left(x+a\right)^2-5^2\)
\(=\left(x+a-5\right)\left(x+a+5\right)\)
d) Ta có: \(\left(x+a\right)^2-\left(y+b\right)^2\)
\(=\left(x+a-y-b\right)\left(x+a+y+b\right)\)
e) Ta có: \(x^2+2x+1-y^2+2y-1\)
\(=\left(x^2+2x+1\right)-\left(y^2-2y+1\right)\)
\(=\left(x+1\right)^2-\left(y-1\right)^2\)
\(=\left(x+1-y+1\right)\left(x+1+y-1\right)\)
\(=\left(x-y+2\right)\left(x+y\right)\)
g) Ta có: \(\left(x^2-2x+1\right)^3+y^6\)
\(=\left[\left(x-1\right)^2\right]^3+y^6\)
\(=\left(x-1\right)^6+y^6\)
\(=\left[\left(x-1\right)^2+y^2\right]\left[\left(x-1\right)^4-\left(x-1\right)^2\cdot y^2+y^4\right]\)
\(=\left(x^2-2x+1+y^2\right)\left(x^4-4x^3+6x^2-4x+1-x^2y^2+2xy^2-y^2+y^4\right)\)
k) Ta có: \(\left(x-a\right)^4+4a^4\)
\(=\left(x-a\right)^4+4a^4+2\cdot\left(x-a\right)^2\cdot2a^2-4\left[a\left(x-a\right)\right]^2\)
\(=\left(x-a+2a^2\right)^2-4\left(ax-a^2\right)^2\)
\(=\left(x-a+2a^2-2ax+2a^2\right)\left(x-a+2a^2+2ax-2a^2\right)\)
\(=\left(x-a-2ax+4a^2\right)\left(x-a+2ax\right)\)
dễ mà cô nương
\(a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)\)
\(\left(a^2+ab+b^2\right)=\left\{\left(a+b\right)^2-ab\right\}\)
\(a^3-b^3=\left(a-b\right)\left(25-6\right)=19\left(a-b\right)\)
ta có
\(a=-5-b\)
suy ra
\(a^3-b^3=19\left(-5-2b\right)\) " xong "
2, trên mạng đầy
3, dytt mọe mày ngu ab=6 thì cmm nó phải chia hết cho 6 chứ :)
4 . \(x^2-\frac{2.1}{2}x+\frac{1}{4}+\frac{1}{3}-\frac{1}{4}>0\) tự làm dcmm
5. trên mạng đầy
6 , trên mang jđầy