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Bài 5:
a. 1 - 2y + y2
= (1 - y)2
b. (x + 1)2 - 25
= (x + 1)2 - 52
= (x + 1 - 5)(x + 1 + 5)
= (x - 4)(x + 6)
c. 1 - 4x2
= 12 - (2x)2
= (1 - 2x)(1 + 2x)
d. 8 - 27x3
= 23 - (3x)3
= (2 - 3x)(4 + 6x + 9x2)
e. (đề hơi khó hiểu ''x3'' !?)
g. x3 + 8y3
= (x + 2y)(x2 - 2xy + y2)
Bài 3:
a:
ĐKXĐ: x>=-5
\(x^2-7x=6\sqrt{x+5}-30\)
=>\(x^2-4x-3x+12=6\sqrt{x+5}-18\)
=>\(\left(x-4\right)\left(x-3\right)=6\left(\sqrt{x+5}-3\right)\)
=>\(\left(x-4\right)\left(x-3\right)=6\cdot\frac{x+5-9}{\sqrt{x+5}+3}\)
=>\(\left(x-4\right)\left(x-3-\frac{6}{\sqrt{x+5}+3}\right)=0\)
=>x-4=0
=>x=4(nhận)
Bài 2:
a: ĐKXĐ: x>=0
\(\sqrt{x+4\sqrt{x}+4}=5x+2\)
=>\(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)
=>\(5x+2=\sqrt{x}+2\)
=>\(5x-\sqrt{x}=0\)
=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)
=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)
b: \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)
=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\) =4
=>|x-1|+|x+2|=4(1)
TH1: x<-2
(1) sẽ trở thành: -x-2+1-x=4
=>-2x-1=4
=>-2x=5
=>x=-5/2(nhận)
TH2: -2<=x<1
(1) sẽ trở thành: x+2+1-x=4
=>3=4(vô lý)
TH3: x>=1
(1) sẽ trở thành: x+2+x-1=4
=>2x+1=4
=>2x=3
=>x=3/2(nhận)
c: ĐKXĐ: x>=1
\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=2\)
=>\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)
=>\(\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=2\)
=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\)
=>\(\left|\sqrt{x-1}-1\right|=2-\sqrt{x-1}-1=1-\sqrt{x-1}\)
=>\(\sqrt{x-1}-1\le0\)
=>\(\sqrt{x-1}\le1\)
=>0<=x-1<=1
=>1<=x<=2
\(a,ĐK:x\ne\pm2\\ A=\dfrac{4x-8+2x+4-5x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\\ ĐK:x\ne-1;x\ne-2\\ B=\dfrac{x+1}{\left(x+1\right)\left(x+2\right)}=\dfrac{1}{x+2}\\ b,x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\ \forall x=0\Leftrightarrow A=\dfrac{1}{0-2}=-\dfrac{1}{2}\\ \forall x=-1\Leftrightarrow A=\dfrac{1}{-1-2}=-\dfrac{1}{3}\)
\(x^2+2x=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=-2\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ \Leftrightarrow B=\dfrac{1}{0+2}=\dfrac{1}{2}\)
1, Với x >= 0 ; x khác 1
\(P=\dfrac{\sqrt{x}\left(x-1\right)+2\sqrt{x}\left(\sqrt{x}-1\right)-\left(3x+1\right)\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}+2x-3\sqrt{x}-3x\sqrt{x}-3x-\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2x\sqrt{x}-x-4\sqrt{x}-1}{\left(x-1\right)\left(\sqrt{x}+1\right)}\)
mình sửa đề câu 2 nhé
a, \(x^2+mx-1=0\)
\(\Delta=m^2-4\left(-1\right)=m^2+4>0\)
Vậy pt luôn có 2 nghiệm pb
b, Theo Vi et : \(\left\{{}\begin{matrix}x_1+x_2=-m\\x_1x_2=-1\end{matrix}\right.\)
Ta có : \(\left(x_1+x_2\right)^2-2x_1x_2=7\)
Thay vào ta được : \(m^2+2=7\Leftrightarrow m^2=5\Leftrightarrow m=\pm\sqrt{5}\)
ĐKXĐ: x∉{3;-3;2;-2}
\(C=\left(\frac{3x-x^2}{9-x^2}-1\right):\left(\frac{9-x^2}{x^2+x-6}-\frac{x-3}{2-x}-\frac{x+2}{x+3}\right)\)
\(=\left(\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}-1\right):\left(\frac{\left(3-x\right)\left(3+x\right)}{\left(x-2\right)\left(x+3\right)}+\frac{x-3}{x-2}-\frac{x+2}{x+3}\right)\)
\(=\left(\frac{x}{x+3}-1\right):\left(\frac{3-x}{x-2}+\frac{x-3}{x-2}-\frac{x+2}{x+3}\right)\)
\(=\frac{x-x-3}{x+3}:\frac{-\left(x+2\right)}{x+3}=\frac{-3}{x+3}\cdot\frac{x+3}{-\left(x+2\right)}=\frac{3}{x+2}\)