Tìm x
(16x-30).8]+80}:5=224
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a. 2x + 70 = 74
<=> 2x = 4
<=> x = 2
b. 120 - \(\dfrac{4x}{2}\) = 80
<=> 120 - 2x = 80
<=> 120 - 80 = 2x
<=> 2x = 40
<=> x = 20
c. (3x + 5)2 = 400
<=> \(|3x+5|=\sqrt{400}\)
<=> \(|3x+5|=20\)
<=> \(\left[{}\begin{matrix}3x+5=20\\3x+5=-20\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-25}{3}\end{matrix}\right.\)
a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)
=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)
=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22
=>-x^2+59x+14-8x^2+5x+22=0
=>-9x^2+54x+36=0
=>x^2-6x-4=0
=>\(x=3\pm\sqrt{13}\)
b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)
=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)
=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32
=>x^2+6x+19=x^2+4x-32
=>2x=-51
=>x=-51/2
a. 2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0
2x=- 4/5 hoặc 3x=1/2
x=-2/5 hoặc x=\(\dfrac{1}{6}\)
b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0
x=2/5 hoặc x=-\(\dfrac{4}{7}\)
d. x(1+5/8-12/16)=1
\(\dfrac{7}{8}\)x=1=> x=8/7
\(156-\left(x+61\right)=82\)
\(\left(x+61\right)=156-82\)
\(\left(x+61\right)=74\)
\(\Rightarrow x=13\)
PP/ss: Lm từng câu ạ ((:
\(156-\left(x+61\right)=82\)
\(\Rightarrow x+61=156-82\)
\(\Rightarrow x+61=74\)
\(\Rightarrow x=74-61\)
\(\Rightarrow x=13\)
\(\left[\left(16x-30\right)\cdot8+80\right]:5=224\\ \Rightarrow2\left(8x-15\right)\cdot8+80=1120\\ \Rightarrow16\left(8x-15\right)=1040\\ \Rightarrow8x+15=65\Rightarrow x=\dfrac{65-15}{8}=\dfrac{25}{4}=6,25\)
\(\dfrac{\left[\left(16x-30\right).8\right]+80}{5}=224\)
<=> \(\dfrac{\left(128x-240\right)+80}{5}=224\)
<=> \(\dfrac{128x-240+80}{5}=224\)
<=> \(\dfrac{128x-160}{5}=224\)
<=> 128x - 160 = 224 . 5
<=> 128x - 160 = 1120
<=> 128x = 1120 +160
<=> 128x = 1280
<=> x = 10