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a) \(3\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)
\(\Leftrightarrow3\sqrt{x}-6\sqrt{x}+4\sqrt{x}=5\)
\(\Leftrightarrow\sqrt{x}=5\)
<=> x = 25
b) pt <=> \(\left(x^2+5\right)=\left(x+1\right)^2\)
<=> \(\left(x^2+5\right)=x^2+2x+1\)
<=> 2x = 4
<=> x = 2
c) pt <=> \(45-14\sqrt{x}+x=x-11\)
<=> \(45+11=14\sqrt{x}\)
<=> \(56=14\sqrt{x}\)
<=> \(4=\sqrt{x}\)
<=> x = 16
p/s : Cậu tự đặt điều kiện nhé
a, \(\sqrt{4-5x}=12\Leftrightarrow4-5x=144\Leftrightarrow5x=140\Leftrightarrow x=28\)
b,ĐK : \(x\ge7\)
\(\sqrt{x^2-14x+49}-3x=1\Leftrightarrow\sqrt{\left(x-7\right)^2}=3x+1\)
\(\Leftrightarrow x-7=3x+1\Leftrightarrow-2x-8=0\Leftrightarrow x=-4\)( vô lí )
c, Bn làm nốt nhé
a) đk: \(x\le\frac{4}{5}\)
Ta có: \(\sqrt{4-5x}=12\)
\(\Leftrightarrow\left|4-5x\right|=144\)
\(\Rightarrow4-5x=144\)
\(\Leftrightarrow5x=-140\)
\(\Rightarrow x=-28\left(tm\right)\)
b) Ta có: \(\sqrt{x^2-14x+49}-3x=1\)
\(\Leftrightarrow\sqrt{\left(x-7\right)^2}=1+3x\)
\(\Leftrightarrow\left|x-7\right|=3x+1\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=3x+1\\x-7=-3x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-8\\4x=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{3}{2}\end{cases}}\)
a, \(\sqrt{16x^2-25}\)
ĐKXĐ : \(16x^2-25\ge0\Leftrightarrow x^2\ge\frac{25}{16}\Leftrightarrow x\le-\frac{5}{4};x\ge\frac{5}{4}\)
b, \(\sqrt{16-9x^2}\)
ĐKXĐ : \(16-9x^2\ge0\Leftrightarrow x^2\le\frac{9}{16}\Leftrightarrow-\frac{3}{4}\le x\le\frac{3}{4}\)
c, \(\sqrt{\frac{x-1}{x+2}}=\frac{\sqrt{x-1}}{\sqrt{x+2}}\)
ĐKXĐ : \(\sqrt{x+2}\ne0\Leftrightarrow x+2\ne0\Leftrightarrow x\ne-2\)
d, \(\frac{1}{\sqrt{x^2-2x-3}}\)
ĐKXĐ : \(\sqrt{x^2-2x-3}\ne0\Leftrightarrow\sqrt{\left(x-1\right)^2-4}\ne0\)
\(\Leftrightarrow\left(x-1-2\right)\left(x-1+2\right)\ne0\Leftrightarrow x\ne-1;3\)
a: \(=\sqrt{5}-1-2\left(\sqrt{2}-1\right)-\left|\sqrt{5}-1-2\left(\sqrt{2}-1\right)\right|\)
\(=\sqrt{5}-1-2\sqrt{2}+2-\left|\sqrt{5}-1-2\sqrt{2}+2\right|\)
\(=-2\sqrt{2}+\sqrt{5}+1-\left(-2\sqrt{2}+\sqrt{5}+1\right)=0\)
b: \(=\left|x-4\right|-\left|x-2\right|\)
\(=\left|3\sqrt{2}-1-4\right|-\left|3\sqrt{2}-1-2\right|\)
\(=\left|3\sqrt{2}-5\right|-\left|3\sqrt{2}-3\right|\)
\(=5-3\sqrt{2}-3\sqrt{2}+3=8-6\sqrt{2}\)
a,4\(\sqrt{x+1}\) -3\(\sqrt{x+1}\) =4 suy ra \(\sqrt{x+1}=4\)suy ra x+1=16 và x=15
b. tương tự phần a suy ra \(5\sqrt{x+1}=\sqrt{x-1}\)suy ra \(^{25\left(x+1\right)=x-1}\)suy ra 24x=-26 suy ra x=\(\frac{-13}{12}\)(ko thỏa mãn đk) nên vô nghiệm
\(a,\sqrt{x+1}=\sqrt{2-x}\)
\(\Rightarrow x+1=2-x\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
a) \(ĐKXĐ:-1\le x\le2\)
Bình phương 2 vế ta có:
\(x+1=2-x\)\(\Leftrightarrow2x=1\)\(\Leftrightarrow x=\frac{1}{2}\)( đpcm )
Vậy \(x=\frac{1}{2}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}+\sqrt{x-1}=16\)
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(\Leftrightarrow2\sqrt{x-1}=16\)\(\Leftrightarrow\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=64\)\(\Leftrightarrow x=65\)( thỏa mãn ĐKXĐ )
Vậy \(x=65\)
c) \(ĐKXĐ:x\ge1\)
\(\sqrt{16x-16}-\sqrt{9x-9}+\sqrt{4x-4}+\sqrt{x-1}=8\)
\(\Leftrightarrow\sqrt{16\left(x-1\right)}-\sqrt{9\left(x-1\right)}+\sqrt{4\left(x-1\right)}+\sqrt{x-1}=8\)
\(\Leftrightarrow4\sqrt{x-1}-3\sqrt{x-1}+2\sqrt{x-1}+\sqrt{x-1}=8\)
\(\Leftrightarrow4\sqrt{x-1}=8\)\(\Leftrightarrow\sqrt{x-1}=2\)
\(\Leftrightarrow x-1=4\)\(\Leftrightarrow x=5\)( thỏa mãn ĐKXĐ )
Vậy \(x=5\)
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