Tìm x:
a) X : 6 = 144
b) 5 x X = 375
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a) Ta có: \(\left|4-5x\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}-5x+4=24\\-5x+4=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x=20\\-5x=-28\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{-4;\dfrac{28}{5}\right\}\)
b) Ta có: \(\left(8+x\right)\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\6-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)
Vậy: \(x\in\left\{-8;6\right\}\)
a) \(\left|4-5x\right|=24\)
\(\Leftrightarrow\left[{}\begin{matrix}4-5x=24\\4-5x=-24\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{28}{5}\end{matrix}\right.\)
b) \(\left(8+x\right)\left(6-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\)
\(a,246:x+34:x=5\\ \Rightarrow\left(246+34\right):x=5\\ \Rightarrow280:x=5\\ \Rightarrow x=56\\ b,360:x-126:x=6\\ \Rightarrow\left(360-126\right):x=6\\ \Rightarrow234:x=6\\ \Rightarrow x=39\)
a: Ta có: \(\left(x-5\right)\left(x+3\right)=x\left(x-3\right)\)
\(\Leftrightarrow x^2-2x-15-x^2+3x=0\)
\(\Leftrightarrow x=15\)
b: Ta có: \(\left(x+2\right)^2=\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow x+2=0\)
hay x=-2
c: Ta có: \(\left(x-6\right)\left(x+6\right)=x^2\)
\(\Leftrightarrow x^2-36=x^2\)(vô lý)
a. (x - 5)(x + 3) = x(x - 3)
<=> x2 + 3x - 5x - 15 = x2 - 3x
<=> x2 - x2 + 3x - 5x + 3x - 15 = 0
<=> x = 15
b. (x + 2)2 = (x - 1)(x + 2)
<=> x2 + 4x + 4 = x2 + 2x - x - 2
<=> x2 - x2 + 4x - 2x + x = -2 - 4
<=> 3x = -5
<=> \(x=\dfrac{-5}{3}\)
c. (x - 6)(x + 6) = x2
<=> x2 - 36 - x2 = 0
<=> x2 - x2 = 36
<=> 0 = 36 (vô lí)
Vậy nghiệm của PT là \(S=\varnothing\)
d. (2x - 3)2 = 4x2 - 8
<=> 4x2 - 12x + 9 - 4x2 + 8 = 0
<=> 4x2 - 4x2 - 12x = -8 - 9
<=> -12x = -17
<=> \(x=\dfrac{17}{12}\)
a) (x + 52400) : 5 = (52400 + 340) : 5
x = 340
b) ( x – 5480) × 6 = (6800 – 5480) × 6
x = 6800.
a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)
\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)
\(\Leftrightarrow2x=-8\)
hay x=-4
b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)
\(\Leftrightarrow-10x=-10\)
hay x=1
c: Ta có: \(\left(2x+3\right)\left(3-2x\right)+\left(2x-1\right)^2=2\)
\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)
\(\Leftrightarrow-4x=-8\)
hay x=2
a: 12:x=144
=>\(x=\dfrac{12}{144}\)
=>\(x=\dfrac{1}{12}\)
b: \(x-17=\left(-2\right)\cdot27\)
=>\(x-17=-54\)
=>\(x=-54+17=-37\)
c: \(3x-125=145\)
=>\(3x=125+145=270\)
=>\(x=\dfrac{270}{3}=90\)
\(a,12:x=144\\ x=\dfrac{12}{144}=\dfrac{1}{12}\\ ---\\ b,x-17=\left(-2\right).27\\ x-17=-54\\ x=-54+17=-37\\ ----\\ 3x-125=145\\ 3x=145+125=270\\ x=\dfrac{270}{3}=90\)
a) Ta có: \(\left(2x-5\right)^3=216\)
\(\Leftrightarrow2x-5=6\)
\(\Leftrightarrow2x=11\)
hay \(x=\dfrac{11}{2}\)
b) Ta có: \(2x-3⋮x+4\)
\(\Leftrightarrow-11⋮x+4\)
\(\Leftrightarrow x+4\in\left\{1;-1;11;-11\right\}\)
hay \(x\in\left\{-3;-5;7;-15\right\}\)
Alo, sugeni two wai phem. Si ga no, you woo be the me that nas te, ai gi da
a) X : 6 = 144
X = 144 * 6
X = 864
b ) 5 * X = 375
X = 375 : 5
X = 75
$ HT & ^_6