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\(a.\sqrt{1-4a+4a^2}-2a=\sqrt{\left(1-2a\right)^2}-2a=\left|1-2a\right|-2a\)
*\(a>\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=2a-1-2a=4a-1\)
* \(a\le\dfrac{1}{2}\Rightarrow\left|1-2a\right|-2a=1-2a-2a=1-4a\)
\(b.x-2y-\sqrt{x^2-4xy+4y^2}=x-2y-\sqrt{\left(x-2y\right)^2}=x-2y-\left|x-2y\right|\)
* \(x\ge2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-x+2y=2x\)
* \(x< 2y\Rightarrow x-2y-\left|x-2y\right|=x-2y-2y+x=2x-4y\)
\(c.x^2+\sqrt{x^4-8x^2+16}=x^2+\sqrt{\left(x^2-4\right)^2}=x^2+\left|x^2-4\right|\)
* \(x^2-4\ge0\Rightarrow x^2+\left|x^2-4\right|=x^2+x^2-4=2x^2-4\)
* \(x^2-4< 0\Rightarrow x^2+\left|x^2-4\right|=x^2+4-x^2=4\)
\(d.2x-1-\dfrac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\dfrac{\sqrt{\left(x-5\right)^2}}{x-5}=2x-1-\dfrac{\left|x-5\right|}{x-5}\)
* \(x\ge5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1-1=2x-2\)
* \(x< 5\Rightarrow2x-1-\dfrac{\left|x-5\right|}{x-5}=2x-1+1=2x\)
\(e.\dfrac{\sqrt{x^4-4x^2+4}}{x^2-2}=\dfrac{\sqrt{\left(x^2-2\right)^2}}{x^2-2}=\dfrac{\left|x^2-2\right|}{x^2-2}\)
* \(x^2\ge2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=1\)
* \(x^2< 2\Rightarrow\dfrac{\left|x^2-2\right|}{x^2-2}=-1\)
\(f.\sqrt{\left(x-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}=\left|x-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}=\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}\)
* \(x\ge4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=x-4+\dfrac{x-4}{x-4}=x-5\)
* \(x< 4\Rightarrow\left|x-4\right|+\dfrac{x-4}{\left|x-4\right|}=4-x-1=5-x\)
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=1-2a-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
Các câu còn lại tương tự nha
\(a,\sqrt{1-4a+4a^2}-2a\)
\(=\sqrt{\left(1-2a\right)^2}-2a\)
\(=\left(1-2a\right)-2a\)
\(=1-4a\)
\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
\(c,x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-2^2\right)^2}\)
\(=x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)
\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)
\(=2x-1-\frac{x-5}{x-5}\)
\(=2x-1-1\)
\(=2x-2\)
\(=2\left(x-1\right)\)
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
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2
a)
\(\sqrt{x^2-2x+1}=x^2-1\)
\(\Rightarrow\left|x-1\right|=x^2-1\)
Với \(x< 1\Rightarrow\left(x-1\right)< 0\)
\(\Rightarrow1-x=x^2-1\)
\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(l\right)\\x=-2\left(tm\right)\end{matrix}\right.\)
Với \(x\ge1\Rightarrow\left(x-1\right)\ge0\)
\(\Rightarrow x-1=x^2-1\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy phương trình đã cho có tập nghiệm \(S=\left\{-2;1\right\}\)
Gõ hết thì mệt lắm , từ câu a --> e làm tương tự còn câu f) bạn bình phương 2 vế lên là được.
a: \(\left\{{}\begin{matrix}x_1+x_2=8\\x_1x_2=6\end{matrix}\right.\)
\(D=x_1^4-x_2^4=\left(x_1+x_2\right)\left(x_1-x_2\right)\left(x_1^2+x_2^2\right)\)
\(=8\cdot\left[\left(x_1+x_2\right)^2-2x_1x_2\right]\cdot\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=8\cdot\left[8^2-2\cdot6\right]\cdot\sqrt{8^2-4\cdot6}\)
\(=8\cdot52\cdot2\sqrt{10}=832\sqrt{10}\)
b: \(E=\left(x_1^2+x_2^2\right)^2-2x_1^2\cdot x_2^2\)
\(=52^2-2\cdot\left(x_1\cdot x_2\right)^2=52^2-2\cdot6^2=2632\)
c: \(F=\dfrac{3x_2^2+3x_1^2}{\left(x_1\cdot x_2\right)^2}=\dfrac{3\cdot52}{6^2}=\dfrac{13}{3}\)