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a) x(x-y)+(x-y)=(x+1)(x-y)
b) 2x+2y -x(x+y)= 2(x+y)-x(x+y)=(2-x)(x+y)
1:=x^3-27-x^2-3=x^3-x^2-30
2: =x-2+125x^3+150x^2+60x+8
=125x^3+150x^2+61x+6
3: \(=2xy-5y+5y=2xy\)
4: =25x-10x^2+15x
=-10x^2+40x
Bài 3:
b: \(\dfrac{1}{x^2-2x}=\dfrac{x+2}{x\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{2}{2x-4}=\dfrac{1}{x-2}=\dfrac{x\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
\(\dfrac{x}{x-2}=\dfrac{x^2\left(x+2\right)}{x\left(x-2\right)\left(x+2\right)}\)
Bài 5:
\(x^2+y^2+1\ge xy+x+y\)
\(\Leftrightarrow2\left(x^2+y^2+1\right)\ge2\left(xy+x+y\right)\)
\(\Leftrightarrow2x^2+2y^2+2\ge2xy+2x+2y\)
\(\Leftrightarrow2x^2+2y^2+2-2xy-2x-2y\ge0\)
\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-2x+1\right)+\left(y^2-2y+1\right)\ge0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-1\right)^2+\left(y-1\right)^2\ge0\left(đúng\right)\)
-Dấu bằng xảy ra \(\Leftrightarrow x=y=1\)
mọi người giải giúp em bài này với ạ em đang cần gấp ạ
a) \(\dfrac{A}{x-2}=\dfrac{x^2+3x+2}{x^2-4}\)
\(\Leftrightarrow\dfrac{A}{x-2}=\dfrac{\left(x+2\right)\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{A}{x-2}=\dfrac{x+1}{x-2}\Leftrightarrow A=x+1\)
b) \(\dfrac{M}{x-1}=\dfrac{x^2+3x+2}{x+1}\)
\(\Leftrightarrow\dfrac{M}{x-1}=\dfrac{\left(x+1\right)\left(x+2\right)}{x+1}\)
\(\Leftrightarrow\dfrac{M}{x-1}=x+2\Leftrightarrow M=\left(x-1\right)\left(x+2\right)=x^2+x-2\)
\(B=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=\left(x^2+x+2\right)^2-16\ge-16\)
Dấu \("="\Leftrightarrow x^2+x+2=0\Leftrightarrow x\in\varnothing\left(x^2+x+2>0\right)\)
Vậy dấu \("="\) ko xảy ra nên sẽ ko tính đc GTNN
Câu 1: A
Câu 2: B
Câu 3: D
Câu 4: A
Câu 5: C
Câu 6: B
Câu 7: A
Câu 9: B
a/
\(\left(x-1\right)^2-\left(x+1\right)^2=2x-6\\ x^2-2x+1-\left(x^2+2x+1\right)=2x-6\\ \)
\(\Leftrightarrow x^2-2x+1-x^2-2x-1-2x+6=0\)
\(\Leftrightarrow6-6x=0\)
=> x=1
cách 1: x2 + 5x + 4
= x2 + x + 4x + 4
= (x2 + x) (4x + 4)
= x(x + 1) + 4(x + 1)
= (x + 1) (x + 4)
cách 2: x2 + 5x + 4
= x2 + 5x + 5 - 1
= (x2 - 1) (5x + 5)
= (x + 1) (x - 1) + 5(x + 1)
= (x + 1) (x - 1 + 5)
= (x + 1) (x + 4)
cách 3: x2 + 5x + 4
= x2 + 2x + 3x + 1 + 3
= (x2 + 2x + 1) (3x + 3)
= (x + 1)2 + 3(x + 1)
= (x + 1) (x + 1 + 3)
= (x + 1) (x + 4)
cách 4: x2 + 5x + 4
= x2 + \(2\frac{5}{2}x+\frac{25}{4}-\frac{9}{4}\)
= \(\left(x+\frac{5}{2}\right)^2-\left(\frac{3}{2}\right)^2\)
= \(\left(x+\frac{5}{2}-\frac{3}{2}\right)\left(x+\frac{5}{2}+\frac{3}{2}\right)\)
= (x + 1) (x + 4)
t i c k nhé!!! 4656576765876898745724724724745756765786876876876876