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Bài 1:
a: \(\Leftrightarrow2-3\sqrt{x}+5\sqrt{x}=8\)
=>2 căn x=6
=>căn x=3
=>x=9
b: \(\Leftrightarrow\dfrac{1}{\sqrt{x}}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{2}{3}:\dfrac{2}{3}=1\)
=>x=1
a) |x| = 4
\(\left[ {_{x = - 4}^{x = 4}} \right.\)
Vậy \(x \in \{ 4; - 4\} \)
b) |x| = \(\sqrt 7 \)
\(\left[ {_{x = - \sqrt 7 }^{x = \sqrt 7 }} \right.\)
Vậy \(x \in \{ \sqrt 7 ; - \sqrt 7 \} \)
c) ) |x+5| = 0
x+5 = 0
x = -5
Vậy x = -5
d) \(\left| {x - \sqrt 2 } \right|\) = 0
x - \(\sqrt 2 \) = 0
x = \(\sqrt 2 \)
Vậy x =\(\sqrt 2 \)
Bài 2:
a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)
Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)
\(\Rightarrow6x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)
\(\Rightarrow4x+12=6x\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\)
Vậy x = 6
b) Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)
\(=\frac{14-5}{8}=\frac{9}{8}\)
+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)
+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)
+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)
Vậy ...
c) \(5^x+5^{x+1}+5^{x+2}=3875\)
\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)
\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)
\(\Rightarrow5^x.31=3875\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
Vậy x = 3
a: \(\Leftrightarrow4x+\dfrac{3}{4}=2\cdot\dfrac{2}{5}+0.01\cdot10=\dfrac{9}{10}\)
=>4x=3/20
hay x=3/80
b: \(\Leftrightarrow\left|x\right|=4+\dfrac{1}{8}-9=-\dfrac{39}{8}\)(vô lý)
c: 2x(x-2/3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
d: \(\dfrac{37-x}{x+13}=\dfrac{3}{7}\)
=>259-7x=3x+39
=>-10x=-220
hay x=22
\(a)\) \(\sqrt{\left(2x-1\right)^2}=3\)
\(\Leftrightarrow\)\(2x-1=3\)
\(\Leftrightarrow\)\(2x=4\)
\(\Leftrightarrow\)\(x=2\)
Vậy \(x=2\)
\(b)\) \(\sqrt{4x^2}=6\)
\(\Leftrightarrow\)\(\sqrt{\left(2x\right)^2}=6\)
\(\Leftrightarrow\)\(2x=6\)
\(\Leftrightarrow\)\(x=3\)
Vậy \(x=3\)
\(c)\) \(\sqrt{4\left(1-x\right)^2}-6=0\)
\(\Leftrightarrow\)\(\sqrt{2^2.\left(1-x\right)^2}=6\)
\(\Leftrightarrow\)\(\sqrt{\left[2\left(1-x\right)\right]^2}=6\)
\(\Leftrightarrow\)\(2-2x=6\)
\(\Leftrightarrow\)\(2x=2-6\)
\(\Leftrightarrow\)\(2x=-4\)
\(\Leftrightarrow\)\(x=-2\)
Vậy \(x=-2\)
\(d)\) \(\sqrt{\left(x-5\right)^2}=\sqrt{\left(3-x\right)^2}\)
\(\Leftrightarrow\)\(x-5=3-x\)
\(\Leftrightarrow\)\(x+x=3+5\)
\(\Leftrightarrow\)\(2x=8\)
\(\Leftrightarrow\)\(x=\frac{8}{2}\)
\(\Leftrightarrow\)\(x=4\)
Vậy \(x=4\)
Chúc bạn học tốt ~
a ) (2x-1)2 =9 => 2x-1 = 3 hoặc 2x -1 = -3
=> 2x=4 hoặc 2x=-2
=> x = 2 hoặc x = -1
b ) 4x2 =36 => x2 = 9 => x = 3 hoặc x = -3
c ) 4x(1-x)2 = 36 => (1-x)2 =9 => 1 - x = 3 hoặc 1 - x = -3
=> x = -2 hoặc x = 4
MinA = 29 \(\Leftrightarrow x=0\)
Min B= 625 \(\Leftrightarrow x=\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
làm rất ngu
ngu như t