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a) \(\sqrt{x^2}=7\)
\(\Leftrightarrow\left|x\right|=7\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)
b) \(\sqrt{\left(x-2020\right)^2}=10\)
\(\Leftrightarrow\left|x-2020\right|=10\)
\(\Leftrightarrow\orbr{\begin{cases}x-2020=10\\x-2020=-10\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2030\\x=2010\end{cases}}\)
c) đk: \(x\ge2\)
\(\sqrt{4}-\left(x-2\right)+3\sqrt{16x-32}=8\)
\(\Leftrightarrow2-x+2+12\sqrt{x-2}=8\)
\(\Leftrightarrow12\sqrt{x-2}=x+4\)
\(\Leftrightarrow144\left(x-2\right)=\left(x+4\right)^2\)
\(\Leftrightarrow x^2-136x+304=0\)
\(\Leftrightarrow\orbr{\begin{cases}x_1=133,726...\\x_2=2,273...\end{cases}}\)
d) đk: \(x\ge-1\)
\(\sqrt{25x+25}-2\sqrt{64x+64}=7\)
\(\Leftrightarrow5\sqrt{x+1}-16\sqrt{x+1}=7\)
\(\Leftrightarrow-11\sqrt{x+1}=7\)
Mà \(-11\sqrt{x+1}\le0< 7\left(\forall x\right)\)
=> pt vô nghiệm
a) \(\sqrt{25x}\) = 35
b) \(\sqrt{4x}\)<= 162
c) \(3\sqrt{x}\) = √12
d) \(2\sqrt{x}\) >=10
a, \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)
\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9\left(x-1\right)}+24\frac{\sqrt{x-1}}{\sqrt{64}}=-17\)
\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+\frac{24\sqrt{x-1}}{8}=-17\)
\(\Rightarrow\frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Rightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\Rightarrow\sqrt{x-1}.-1=-17\)
\(\Rightarrow\sqrt{x-1}=17\)
\(\Rightarrow x-1=289\)
\(\Rightarrow x=290\)
b, \(3x-7\sqrt{x}+4=0\)
\(\Rightarrow3x-3\sqrt{x}-4\sqrt{x}+4=0\)
\(\Rightarrow3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=0\)
\(\Rightarrow\left(\sqrt{x}-1\right)\left(3\sqrt{x}-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}-1=0\\3\sqrt{x}-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}\sqrt{x}=1\\3\sqrt{x}=4\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{16}{9}\end{cases}}}\)
c, \(-5x+7\sqrt{x}+12=0\)
\(\Rightarrow-5x-5\sqrt{x}+12\sqrt{x}+12=0\)
\(\Rightarrow-5\sqrt{x}\left(\sqrt{x}+1\right)+12\left(x+1\right)=0\)
\(\Rightarrow\left(\sqrt{x}+1\right)\left(-5\sqrt{x}+12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x}+1=0\\-5\sqrt{x}+12=0\end{cases}\Rightarrow\orbr{\begin{cases}\sqrt{x}=-1VN\\-5\sqrt{x}=-12\end{cases}}\Rightarrow\orbr{\begin{cases}\\\sqrt{x}=\frac{12}{5}\end{cases}\Rightarrow}\orbr{\begin{cases}\\x=\frac{144}{25}\end{cases}}}\)
1) ĐK: \(x-1\ge0\Leftrightarrow x\ge1\)
pt \(\Leftrightarrow\frac{1}{2}\sqrt{x-1}-\frac{3}{2}.3\sqrt{x-1}+\frac{24}{8}\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}\left(\frac{1}{2}-\frac{9}{2}+3\right)=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=17^2=289\Leftrightarrow x=290\left(tm\right)\)
b) \(3x-7\sqrt{x}+4=0\)
ĐK: \(x\ge0\)
Đặt \(\sqrt{x}=t\left(t\ge0\right)\Leftrightarrow t^2=x\)
Ta có phương trình ẩn t:
\(3t^2-7t+4=0\)( giải đen ta)
\(\Leftrightarrow\orbr{\begin{cases}t=1\\t=\frac{4}{3}\end{cases}}\)
Với t=1 ta có: \(\sqrt{x}=1\Leftrightarrow x=1\) (tm)
Với t=4/3 ta có: \(\sqrt{x}=\frac{4}{3}\Leftrightarrow x=\frac{16}{9}\) (tm)
Câu c em làm tương tự câu b nhé!
a) điều kiện : \(x-5\ge0\Leftrightarrow x\ge5\)
\(\sqrt{x-5}=3\Leftrightarrow\left(\sqrt{x-5}\right)^2=3^2\Leftrightarrow\left|x-5\right|=9\Leftrightarrow x-5=9\)
\(\Leftrightarrow x=9+5\Leftrightarrow x=14\) vậy \(x=14\)
b) điều kiện : \(x-10\ge0\Leftrightarrow x\ge10\)
\(\sqrt{x-10}=-2\) ta có : \(\sqrt{x-10}\ge0\) với mọi \(x\)
\(\Rightarrow\sqrt{x-10}=-2\) là vô nghiệm
c) điều kiện : \(2x-1\ge0\Leftrightarrow2x\ge1\Leftrightarrow x\ge\dfrac{1}{2}\)
\(\sqrt{2x-1}=\sqrt{5}\Rightarrow2x-1=5\Leftrightarrow2x=5+1\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}\Leftrightarrow x=3\) vậy \(x=3\)
d) điều kiện : \(4-5x\ge0\Leftrightarrow5x\le4\Leftrightarrow x\le\dfrac{4}{5}\)
\(\sqrt{4-5x}=12\Leftrightarrow\left(\sqrt{4-5x}\right)^2=12^2\Leftrightarrow\left|4-5x\right|=144\)
\(\Leftrightarrow4-5x=144\Leftrightarrow-5x=144-4\Leftrightarrow-5x=140\Leftrightarrow x=\dfrac{140}{-5}\)
\(\Leftrightarrow x=-28\) vậy \(x=-28\)
Hướng dẫn giải:
a) ĐS: -√3. b) ĐS: 3434(a - 3).
c) √9+12a+4a2b29+12a+4a2b2 = √(3+2a)2b2(3+2a)2b2 = √(3+2a)2b2(3+2a)2b2 = |3+2a||b||3+2a||b|
Vì b < 0 nên |b||b| = -b.
Vì a > -1,5 nên 3 + 2a > 0. Do đó = 3+ 2a.
Vậy √9+12a+4a2b29+12a+4a2b2 = -3+2ab3+2ab.
d) ĐS: -√ab
a: \(x^2=121\)
=>\(\left[{}\begin{matrix}x=11\\x=-11\end{matrix}\right.\)
b: \(x^2=64\)
=>\(\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
c: ĐKXĐ: x>=1
\(2\sqrt{x-1}=12\)
=>\(\sqrt{x-1}=12:2=6\)
=>x-1=36
=>x=36+1=37(nhận)
d: \(5\cdot\sqrt[3]{x+3}=10\)
=>\(\sqrt[3]{x+3}=\dfrac{10}{5}=2\)
=>x+3=8
=>x=8-3=5
a) x=√121. x=11. b) x=√64. x=8.