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Câu a:
|\(\sqrt2\) - \(x\)| = \(\sqrt2\)
\(\left[\begin{array}{l}\sqrt2-x=\sqrt2\\ \sqrt2-x=-\sqrt2\end{array}\right.\)
\(\left[\begin{array}{l}x=0\\ x=2\sqrt2\end{array}\right.\)
Vậy \(x\in\) {0; \(2\sqrt2\)}
Câu b:
|\(x-1\)| = \(\sqrt3\) + 2
\(\left[\begin{array}{l}x-1=\sqrt3+2\\ x-1=-\sqrt{3-2}\end{array}\right.\)
\(\left[\begin{array}{l}x=\sqrt3+2+1\\ x=-\sqrt3-2+1\end{array}\right.\)
\(\left[\begin{array}{l}x=\sqrt3+\left(2+1\right)\\ x=-\sqrt3-\left(2-1\right)\end{array}\right.\)
\(\left[\begin{array}{l}x=\sqrt3+3\\ x=-\sqrt3-1\end{array}\right.\)
Vậy \(x\in\) {- \(\sqrt3\) - 1; \(\sqrt3\) + 3}
\(\sqrt{\left(x-\sqrt{2}\right)^2};\sqrt{\left(y+\sqrt{2}\right)};lx+y+zl\ge0\Rightarrow\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{\left(y+\sqrt{2}\right)^2}=lx+y+zl=0\)
\(\Rightarrow x-\sqrt{2}=y+\sqrt{2}=x+y+z=0\Rightarrow x=\sqrt{2};y=-\sqrt{2}\Rightarrow z=0\)
vậy (x;y;z)=\(\left(\sqrt{2};-\sqrt{2};0\right)\)
Nhận xét: \(\sqrt{\left(x-\sqrt{2}\right)^2}\ge0;\sqrt{\left(y+\sqrt{2}\right)^2}\ge0;\left|x+y+z\right|\ge0\)
Để \(\sqrt{\left(x-\sqrt{2}\right)^2}+\sqrt{\left(y+\sqrt{2}\right)^2}+\left|x+y+z\right|=0\)thì
\(\sqrt{\left(x-\sqrt{2}\right)^2}=\sqrt{\left(y+\sqrt{2}\right)^2}=\left|x+y+z\right|=0\)
=> \(x-\sqrt{2}=0;y+\sqrt{2}=0;x+y+z=0\)
=> \(x=\sqrt{2};y=-\sqrt{2};z=-x-y=0\)
Vậy...
a) \(\sqrt{x}=2\)
\(\Rightarrow x=2^2=4\)
b) \(\sqrt{x-1-5}=\sqrt{x-6}=0\)
\(\Rightarrow x-6=0^2=0\)
\(\Rightarrow x=6\)
các câu sau tương tự
a) \(-0,6^0+\frac{1}{2}.2-3x=-\frac{1}{4}\)
\(\Leftrightarrow-1+1-3x=-\frac{1}{4}\Leftrightarrow-3x=-\frac{1}{4}\Leftrightarrow3x=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}:3=\frac{1}{12}\)
b)\(2^{x-2}+22=3.2^x\Leftrightarrow3.2^x-2^{x-2}=22\Leftrightarrow2^{x-2}\left(3.2^2-1\right)=22\)
\(\Leftrightarrow2^{x-2}.11=22\Leftrightarrow2^{x-2}=2\Leftrightarrow x-2=1\Leftrightarrow x=3\)
c) \(\left(x-1\right)^2=\sqrt{\left(-\frac{9}{16}\right)^2}\Leftrightarrow\left(x-1\right)^2=\frac{9}{16}\Leftrightarrow\left(x-1\right)^2=\left(\frac{3}{4}\right)^2\)
TH1: x - 1 = 3/4 => x = 3/4 + 1 => x = 7/4
Th2: x - 1 = - 3/4 => x = -3/4 +1 => x = 1/4
d) \(\Leftrightarrow\sqrt{x^2+2}=12-5=7\Leftrightarrow x^2+2=7^2\Leftrightarrow x^2=49-2\Leftrightarrow x^2=47\)
\(x=\sqrt{47};x=-\sqrt{47}\)
<=> \(4\sqrt{x}-13=16-5\)
\(4\sqrt{x}=11+13\)
\(\sqrt{x}=24:4\)
\(\sqrt{x}=6\)
\(x=36\)
Ta có:
4√x − √169 = √256 − (√5)^2
⇔ 4√x − 13 = 16 − 5
⇔ 4√x − 13 = 11
⇔ 4√x = 24
⇔ √x = 6
⇔ x = 36
Vì x = 36 ≥ 0 nên thỏa mãn điều kiện.
Đáp án: x = 36.
X=0,1