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a, \(-4x+5+2x-1=3\Leftrightarrow-2x=-1\Leftrightarrow x=\dfrac{1}{2}\)
b, \(-2x+2=2\Leftrightarrow x=0\)
c, \(-2x-6=-8\Leftrightarrow x=1\)
a)\(\left|\dfrac{x-1}{3}\right|=\dfrac{11}{5}\Rightarrow\dfrac{x-1}{3}=\pm\dfrac{11}{5}\\ \Rightarrow\left[{}\begin{matrix}\dfrac{x-1}{3}=\dfrac{11}{5}\\\dfrac{x-1}{3}=-\dfrac{11}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{33}{5}\\x-1=\dfrac{-33}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{38}{5}\\x=\dfrac{-28}{5}\end{matrix}\right.\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
Bạn nên viết lại đề bài cho sáng sủa, rõ ràng để người đọc dễ hiểu hơn.
f: =>4(x^2+4x-5)-x^2-7x-10=3(x^2+x-2)
=>4x^2+16x-20-x^2-7x-10-3x^2-3x+6=0
=>6x-24=0
=>x=4
e: =>8x+16-5x^2-10x+4(x^2-x-2)=4-x^2
=>-5x^2-2x+16+4x^2-4x-8=4-x^2
=>-6x+8=4
=>-6x=-4
=>x=2/3
d: =>2x^2+3x^2-3=5x^2+5x
=>5x=-3
=>x=-3/5
b: =>2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20
=>-12x-2=-17x+20
=>5x=22
=>x=22/5
b) Ta có: \(-5+\left|3x-1\right|+6=\left|-4\right|\)
\(\Leftrightarrow\left|3x+1\right|+1=4\)
\(\Leftrightarrow\left|3x+1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=3\\3x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{3};-\dfrac{4}{3}\right\}\)
c) Ta có: \(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Leftrightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Leftrightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-1-1\right)\left(x-1+1\right)=0\)
\(\Leftrightarrow x\cdot\left(x-1\right)^2\cdot\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;1;2\right\}\)
d) Ta có: \(5^{-1}\cdot25^x=125\)
\(\Leftrightarrow5^{-1}\cdot5^{2x}=5^3\)
\(\Leftrightarrow5^{2x-1}=5^3\)
\(\Leftrightarrow2x-1=3\)
\(\Leftrightarrow2x=4\)
hay x=2
Vậy: x=2
Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
