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Cho biểu thức
c=(căng x-2/căng x+2+căng x+2/căng x-2)nhân căng x+2/2 - 4 căng x/căng x-2
a)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{a-9}\)
\(P=\frac{\sqrt{a}}{\sqrt{a}+3}+\frac{2\sqrt{a}}{\sqrt{a}-3}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{\sqrt{a}\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}+\frac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\frac{3a+9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{a-3\sqrt{a}+3+3\sqrt{a}-3a-9}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}\)
\(P=\frac{-2a-3}{a-9}\)
b) Để \(P=\frac{1}{3}\Rightarrow\frac{-2a-3}{a-9}=\frac{1}{3}\)
\(\Rightarrow3\left(-2a-3\right)=a-9\)
\(\Rightarrow-6a-9=a-9\)
\(\Rightarrow-6a-a=-9+9\)
\(\Rightarrow-7a=0\left(L\right)\)
Vậy ko có gt của a để P=1/3 ( mk ko chắc.....)
a)\(x+3+\sqrt{x^2-6x+9}\)
\(=x+3+\sqrt{\left(x-3\right)^2}\)
\(=x+3+x-3\)
\(=2x\)
b)\(\sqrt{x^2+4x+4}-\sqrt{x^2}\)
\(=\sqrt{\left(x+2\right)^2}-x\)
\(=x+2-x\)
=2
c)\(\sqrt{\frac{x^2-2x+1}{x-1}}\)
\(=\sqrt{\frac{\left(x-1\right)^2}{x-1}}\)
\(=\sqrt{x-1}\)
k) ĐK: $x^2\geq 5$
PT $\Leftrightarrow 2\sqrt{x^2-5}-\frac{1}{3}\sqrt{x^2-5}+\frac{3}{4}\sqrt{x^2-5}-\frac{5}{12}\sqrt{x^2-5}=4$
$\Leftrightarrow 2\sqrt{x^2-5}=4$
$\Leftrightarrow \sqrt{x^2-5}=2$
$\Rightarrow x^2-5=4$
$\Leftrightarrow x^2=9\Rightarrow x=\pm 3$ (đều thỏa mãn)
l) ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 2\sqrt{x+1}+3\sqrt{x+1}-\sqrt{x+1}=4$
$\Leftrightarrow 4\sqrt{x+1}=4$
$\Leftrightarrow \sqrt{x+1}=1$
$\Rightarrow x+1=1$
$\Rightarrow x=0$
m)
ĐKXĐ: $x\geq -1$
PT $\Leftrightarrow 4\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}+3\sqrt{x+1}$
$\Leftrightarrow 6\sqrt{x+1}=16+2\sqrt{x+1}$
$\Leftrightarrow 4\sqrt{x+1}=16$
$\Leftrightarrow \sqrt{x+1}=4$
$\Rightarrow x=15$ (thỏa mãn)
h)
ĐKXĐ: $x\geq -5$
PT $\Leftrightarrow \sqrt{x+5}=6$
$\Rightarrow x+5=36\Rightarrow x=31$ (thỏa mãn)
i) ĐKXĐ: $x\geq 5$
PT \(\Leftrightarrow \sqrt{x-5}+4\sqrt{x-5}-\sqrt{x-5}=12\)
\(\Leftrightarrow 4\sqrt{x-5}=12\Leftrightarrow \sqrt{x-5}=3\Rightarrow x-5=9\Rightarrow x=14\) (thỏa mãn)
j)
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow 3\sqrt{2x}+\sqrt{2x}-6\sqrt{2x}+4=0$
$\Leftrightarrow -2\sqrt{2x}+4=0$
$\Leftrightarrow \sqrt{2x}=2$
$\Rightarrow x=2$ (thỏa mãn)
a) \(2x-\sqrt{4x^2+4x+1}=2x-\sqrt{\left(2x+1\right)^2}=2x-\left|2x+1\right|\)
Vì \(x< -\frac{1}{2}\)nên \(\left|2x+1\right|=-\left(2x+1\right)\)
\(\Rightarrow2x+2x+1=4x+1\)
b) \(3x+2-\sqrt{9x^2-12x+4}=3x+2-\sqrt{\left(3x-2\right)^2}=3x+2-\left|3x-2\right|\)
Khi \(x\ge\frac{2}{3}\)thì \(\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-3x+2=4\)
Khi \(x< \frac{2}{3}\) thì \(\left|3x-2\right|=2-3x\)
\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-\left(2-3x\right)=6x\)
c) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)
Đặt \(\sqrt{a}=x\) ta được : \(3x-4x+7x=6x\)\(=6\sqrt{a}\)( Do \(a\ge0\))
d) \(\sqrt{160a}+2\sqrt{40a}-3\sqrt{90a}=4\sqrt{10a}+4\sqrt{10a}-9\sqrt{10a}\)\(=-\sqrt{10}\)
TK NKA !!!
1) Ta có: \(\left(\sqrt{12}-6\sqrt{3}+\sqrt{24}\right)\cdot\sqrt{6}-\left(\frac{5}{2}\sqrt{2}+12\right)\)
\(=\left(2\sqrt{3}-6\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{25}{4}\cdot2}+12\right)\)
\(=\left(-4\sqrt{3}+2\sqrt{6}\right)\cdot\sqrt{6}-\left(\sqrt{\frac{50}{4}}+12\right)\)
\(=-12\sqrt{2}+12-\frac{5\sqrt{2}}{2}-12\)
\(=\frac{-24\sqrt{2}-5\sqrt{2}}{2}\)
\(=\frac{-29\sqrt{2}}{2}\)
2) Ta có: \(\frac{26}{2\sqrt{3}+5}-\frac{4}{\sqrt{3}-2}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{\left(5+2\sqrt{3}\right)\left(5-2\sqrt{3}\right)}+\frac{4}{2-\sqrt{3}}\)
\(=\frac{26\left(5-2\sqrt{3}\right)}{25-12}+\frac{4\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=2\left(5-2\sqrt{3}\right)+4\left(2+\sqrt{3}\right)\)
\(=10-4\sqrt{3}+8+4\sqrt{3}\)
\(=18\)
3) ĐK để phương trình có nghiệm là: x≥0
Ta có: \(\sqrt{x^2-6x+9}=2x\)
\(\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x\)
\(\Leftrightarrow\left|x-3\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x\\x-3=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3-2x=0\\x-3+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x-3=0\\3x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=3\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1}
4) ĐK để phương trình có nghiệm là: \(x\ge\frac{1}{2}\)
Ta có: \(\sqrt{4x^2+1}=2x-1\)
\(\Leftrightarrow\left(\sqrt{4x^2+1}\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow4x^2+1=4x^2-4x+1\)
\(\Leftrightarrow4x^2+1-4x^2+4x-1=0\)
\(\Leftrightarrow4x=0\)
hay x=0(loại)
Vậy: S=∅
Câu 1: Sửa lạ đề chút nhé : 4x + 1 -> 4x -1
Đặt A = \(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)
=> \(\sqrt{2}.A\)= \(\sqrt{4x-1+2\sqrt{4x-1}+1}+\sqrt{4x-1-2\sqrt{4x-1}+1}\)
= \(\sqrt{\left(\sqrt{4x-1}+1\right)^2}+\sqrt{\left(\sqrt{4x-1}-1\right)^2}\)
= \(\left|\sqrt{4x-1}+1\right|+\left|\sqrt{4x-1}-1\right|\)
Vì \(\frac{1}{4}< x< \frac{1}{2}\Rightarrow0< 4x-1< 1\Rightarrow0< \sqrt{4x-1}< 1\)
nên \(\sqrt{2}A=\)\(\sqrt{4x-1}+1+1-\sqrt{4x-1}\)=2
=> \(A=2:\sqrt{2}=\sqrt{2}\)
Câu 2. Có: \(9-4\sqrt{2}=8-2.2\sqrt{2}+1=\left(2\sqrt{2}-1\right)^2\)
=> \(\sqrt{9-4\sqrt{2}}=2\sqrt{2}-1\)
=> \(4+\sqrt{9-4\sqrt{2}}=4+2\sqrt{2}-1=2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)
=> \(\sqrt{4+\sqrt{9-4\sqrt{2}}}=\sqrt{2}+1\)
=> \(53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}=53-20\left(\sqrt{2}+1\right)=33-2.10\sqrt{2}=5^2-2.5.2\sqrt{2}+8=\left(5-2\sqrt{2}\right)^2\)
=> \(\sqrt{53-20\sqrt{4+\sqrt{9-4\sqrt{2}}}}=5-2\sqrt{2}\)
\(\sqrt{2x+\sqrt{4x-1}}+\sqrt{2x-\sqrt{4x-1}}\)