3(x+2)-2(x+1)=26
Tìm x
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\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15};\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\\ \Rightarrow\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x+y-z}{10+15-12}=\dfrac{-26}{13}=-2\\ \Rightarrow\left\{{}\begin{matrix}x=-20\\y=-30\\z=-24\end{matrix}\right.\)
a: =>9x^2+12x+4-9x^2+12x-4=5x+38
=>24x=5x+38
=>19x=38
=>x=2
e: =>x^3+1-2x=x^3-x
=>-2x+1=-x
=>-x=-1
=>x=1
f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1
=>12x-9=3x+1
=>9x=10
=>x=10/9
b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
=>-3x+3=3x-9
=>-6x=-12
=>x=2
1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)
\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
a. (x - 2)(x + 2) - (x - 3)2 = 9
<=> x2 - 22 - (x - 3)2 = 32
<=> x - 2 - (x - 3) = 3
<=> x - 2 - x + 3 = 3
<=> x - x = 3 - 3 + 2
<=> 0 = 2 (Vô lí)
Vậy nghiệm của PT là S = \(\varnothing\)
b: Ta có: \(\left(x-1\right)\left(x^2+1\right)-\left(x+1\right)\left(x^2-x+1\right)=x\left(2-x\right)\)
\(\Leftrightarrow x^3+x-x^2-1-x^3-1=2x-x^2\)
\(\Leftrightarrow-x^2+x-2-2x+x^2=0\)
\(\Leftrightarrow-x=2\)
hay x=-2
3(x+2)-2(x+1)=26
3(x+1)+3-2(x+1)=26
(3-2).(x+1)+3=26
x+1=26-3
x+1=23
x=23-1
x=22
Vậy x=22