Tìm x
( 2^6 - 4x ) . 8^39 = 2 . 8^40
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a: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{9\left(x-2\right)^2}=18\)
=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)
=>\(3\cdot\left|x-2\right|=18\)
=>\(\left|x-2\right|=6\)
=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
=>\(4\sqrt{x-2}=40\)
=>\(\sqrt{x-2}=10\)
=>x-2=100
=>x=102(nhận)
d: ĐKXĐ: \(x\in R\)
\(\sqrt{4\left(x-3\right)^2}=8\)
=>\(\sqrt{\left(2x-6\right)^2}=8\)
=>|2x-6|=8
=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: \(x\in R\)
\(\sqrt{4x^2+12x+9}=5\)
=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)
=>\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
f: ĐKXĐ:x>=6/5
\(\sqrt{5x-6}-3=0\)
=>\(\sqrt{5x-6}=3\)
=>\(5x-6=3^2=9\)
=>5x=6+9=15
=>x=15/5=3(nhận)
a) \(55-4x-4\left(-x+3\right)=6-2\left(-8-3x\right)\)
\(55-4x+4x-12=6+16+6x\)
\(43-6-16=6x\)
\(6x=21\)
\(x=3,5\)
b) \(-5\left(-2x-6\right)-9\left(4-7x\right)=51-3x+6\left(x-9\right)\)
\(10x+30-36+63x=51-3x+6x-54\)
\(73x-6=-3+3x\)
\(73x-3x=-3+6\)
\(70x=3\)
\(x=\frac{3}{70}\)
c) \(93+\left|6-3x\right|-39=231\)
\(\left|6-3x\right|+54=231\)
\(\left|6-3x\right|=177\)
\(\Rightarrow\orbr{\begin{cases}6-3x=177\\6-3x=-177\end{cases}}\Rightarrow\orbr{\begin{cases}3x=6-177\\3x=6+177\end{cases}}\Rightarrow\orbr{\begin{cases}3x=-171\\3x=183\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-57\\x=61\end{cases}}\)
\(\left(4x-8\right)\left(3x-6\right)>\left(4x-8\right)\left(2x+2\right)\)
\(\Leftrightarrow\)\(\left(4x-8\right)\left(3x-6\right)-\left(4x-8\right)\left(2x+2\right)>0\)
\(\Leftrightarrow\)\(\left(4x-8\right)\left(3x-6-2x-2\right)>0\)
\(\Leftrightarrow\)\(4\left(x-2\right)\left(x-8\right)>0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x-2>0\\x-8>0\end{cases}}\)hoặc \(\hept{\begin{cases}x-2< 0\\x-8< 0\end{cases}}\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x>2\\x>8\end{cases}}\)hoặc \(\hept{\begin{cases}x< 2\\x< 8\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x>8\\x< 2\end{cases}}\)
Vậy...
c) (4x - 8)[x + (-3)] = 0
=> 4(x - 2)(x - 3) = 0
=> (x - 2)(x - 3) = 0
=> x - 2 = 0 hoặc x - 3 = 0
+) x - 2 = 0 => x = 2
+) x - 3 = 0 => x = 3
Vậy x \(\in\){2;3}
11(x - 6) = 4x + 11
=> 11x - 66 = 4x + 11
=> 11x - 4x = 11 + 66
=> 7x = 77
=> x = 77/7
=> x = 11
( 26 - 4x ) . 839 = 2 . 840
26 - 4x = 2 . 840 : 839
26 - 4x = 2 . 8
4x = 26 - 16
4x = 64 - 16
4x = 48
x = 48 : 4
x = 12
( 2^6 - 4x ) . 8 ^39 = 2. 8 ^ 40
( 2 ^6 - 4x ) . 8 ^ 39 = 2 . 8 . 8 ^ 39
( 2 ^6 - 4x ) = 16
64 - 4x = 16
4x = 64 - 16
4x = 48
x = 48 : 4
x = 12