Tìm x biết
a)|x-1|+|x-3|=4
b) |x^2+|6x-2||=x^2+4
c) |2x+3|-2|4-x|=5
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\(a,\text{Với }x< -2\Rightarrow3-x-x-2=4\\ \Rightarrow-2x=3\Rightarrow x=-\dfrac{3}{2}\left(ktm\right)\\ \text{Với }-2\le x< 3\Rightarrow3-x+x+2=4\\ \Rightarrow0x=-1\Rightarrow x\in\varnothing\\ \text{Với }x\ge3\Rightarrow x-3+x+2=4\\ \Rightarrow2x=5\Rightarrow x=\dfrac{5}{2}\left(ktm\right)\)
Vậy \(x\in\varnothing\)
\(b,\text{Với }x< 2\Rightarrow4-2x+18-6x=21\\ \Rightarrow22-8x=21\Rightarrow x=\dfrac{1}{8}\left(tm\right)\\ \text{Với }2\le x< 3\Rightarrow2x-4+18-6x=21\\ \Rightarrow-4x+14=21\Rightarrow x=-\dfrac{7}{4}\left(ktm\right)\\ \text{Với }x\ge3\Rightarrow2x-4+6x-18=21\\ \Rightarrow8x=43\Rightarrow x=\dfrac{43}{8}\left(tm\right)\)
Vậy \(x\in\left\{\dfrac{1}{8};\dfrac{43}{8}\right\}\)
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
f: Ta có: \(x^3-6x^2+12x-19=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)
\(\Leftrightarrow\left(x-2\right)^3=11\)
hay \(x=\sqrt[3]{11}+2\)
a: \(\left|3x-2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)
b: =>4x^2+8x-8x^2+5x-10=0
=>-4x^2+13x-10=0
=>x=2 hoặc x=5/4
c: =>2x^2-5x+6x-15=2x^2+8x
=>x-15=8x
=>-7x=15
=>x=-15/7
d: =>3x^2+15x-2x-10-3x^2-12x=5
=>x-10=5
=>x=15
e: =>x^2-3x+2x^2+2x=3x^2-12
=>-x=-12
=>x=12
v) \(\left(-\dfrac{1}{2}x+3\right)\left(2x+6-4c^3\right)\)
\(=-\dfrac{1}{2}\left(2x+6-4c^3\right)+3\left(2x+6-4c^3\right)\)
\(=-x^2-3x+2c^3x+6x+18-12c^3\)
\(=-x^2+3x+2c^3x+18-12c^3\)
f) \(\left(2x-5\right)\left(x^2-x+3\right)\)
\(=2x\left(x^2-x+3\right)-5\left(x^2-x+3\right)\)
\(=2x^3-2x^2+6x-5x^2+5x-15\)
\(=2x^3-7x^2+11x-15\)
w) \(\left(3x+1\right)\left(x^2-2x-5\right)\)
\(=3x\left(x^2-2x-5\right)+\left(x^2-2x-5\right)\)
\(=3x^3-6x^2-15x+x^2-2x-5\)
\(=3x^3-5x^2-17x-5\)
x) \(\left(6x-3\right)\left(x^2+x-1\right)\)
\(=6x\left(x^2+x-1\right)-3\left(x^2+x-1\right)\)
\(=6x^3+6x^2-6x-3x^2-3x+3\)
\(=6x^3+3x^2-9x+3\)
y) \(\left(5x-2\right)\left(3x+1-x^2\right)\)
\(=5x\left(3x+1-x^2\right)-2\left(3x+1-x^2\right)\)
\(=15x^2+5x-5x^3-6x-2+2x^2\)
\(=-5x^3+17x^2-x-2\)
z) \(\left(\dfrac{3}{4}x+1\right)\left(4x^2+4x+4\right)\)
\(=\dfrac{3}{4}x\left(4x^2+4x+4\right)+\left(4x^2+4x+4\right)\)
\(=3x^3+3x^2+3x+4x^2+4x+4\)
\(=3x^3+7x^2+7x+4\)
f: =2x^3-2x^2+6x-5x^2+5x-15
=2x^3-7x^2+11x-15
w: =3x^3-6x^2-15x+x^2-2x-5
=3x^3-5x^2-17x-5
x: =6x^3+6x^2-6x-3x^2-3x+3
=6x^3+3x^2-9x+3
y: =(5x-2)(-x^2+3x+1)
=-5x^3+15x^2+5x+2x^2-6x-2
=-5x^3+17x^2-x-2
z: =3x^3+3x^2+3x+4x^2+4x+4
=3x^3+7x^2+7x+4
\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)
\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)
\(\Rightarrow27x+15=96\)
\(\Rightarrow27x=81\)
\(\Rightarrow x=3\left(tm\right)\)
\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\left(tm\right)\)
#Toru
a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)
\(\Rightarrow-6x+8x+3x+3+4x+2=32\)
\(\Rightarrow9x+5=32\)
\(\Rightarrow9x=32-5\)
\(\Rightarrow9x=27\)
\(\Rightarrow x=\dfrac{27}{9}\)
\(\Rightarrow x=3\)
b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\))
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=\dfrac{12}{2}\)
\(\Rightarrow x=6\left(tm\right)\)
a: =>2x-x=-5/2-1/3
=>x=-17/6
b: =>4(x-2)2=36
=>(x-2)2=9
=>x-2=3 hoặc x-2=-3
hay x=5 hoặc x=-1
c: =>2x+1/2=5/6
=>2x=1/3
hay x=1/6
a) ĐKXĐ: x≠-5
Ta có: \(\dfrac{2x-5}{x+5}=4\)
\(\Leftrightarrow2x-5=4\left(x+5\right)\)
\(\Leftrightarrow2x-5=4x+20\)
\(\Leftrightarrow2x-5-4x-20=0\)
\(\Leftrightarrow-2x-25=0\)
\(\Leftrightarrow-2x=25\)
hay \(x=\dfrac{-25}{2}\)(nhận)
Vậy: \(S=\left\{-\dfrac{25}{2}\right\}\)
b) ĐKXĐ: x≠0
Ta có: \(\dfrac{x^2-4}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-4\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-8=2x^2+3x\)
\(\Leftrightarrow2x^2-8-2x^2-3x=0\)
\(\Leftrightarrow-3x-8=0\)
\(\Leftrightarrow-3x=8\)
hay \(x=\dfrac{-8}{3}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{3}\right\}\)
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-5\right\}\)
Ta có: \(\dfrac{2x+3}{2x-1}=\dfrac{x-3}{x+5}\)
\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)
\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)
\(\Leftrightarrow2x^2+13x+15-2x^2+7x-3=0\)
\(\Leftrightarrow20x+12=0\)
\(\Leftrightarrow20x=-12\)
hay \(x=-\dfrac{3}{5}\)(nhận)
Vậy: \(S=\left\{-\dfrac{3}{5}\right\}\)
d) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(x+7\right)\left(6x+1\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+x+42x+7\)
\(\Leftrightarrow6x^2-13x+6=6x^2+43x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)