1) Tim x
a) x + 1/2 . x - 25% . x = 10
b) x + x -1 + x-2 + x-3 + x-4 +........+x - 5= 255
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a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)
\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)
\(\Rightarrow6x^2+2-6x^2+12x-6=10\)
\(\Rightarrow12x-4=10\)
\(\Rightarrow12x=14\)
\(\Rightarrow x=\dfrac{7}{6}\)
b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Rightarrow x^3-25x-x^3-8=42\)
\(\Rightarrow-25x-8=42\)
\(\Rightarrow-25x=50\)
\(\Rightarrow x=\dfrac{50}{-25}=-2\)
c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Rightarrow24x+25=49\)
\(\Rightarrow24x=24\)
\(\Rightarrow x=\dfrac{24}{24}=1\)
\(a,\Leftrightarrow4x^2-24x+36-4x^2+1=10\\ \Leftrightarrow-24x=-27\Leftrightarrow x=\dfrac{9}{8}\\ b,\Leftrightarrow x\left(x^2-25\right)=0\\ \Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
\(a,4.\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)
\(\Leftrightarrow4.\left(x^2-6x+9\right)-\left(2x^2\right)-1^2=10\)
\(\Leftrightarrow4x^2-24x+36-4x^2+1=10\)
\(\Leftrightarrow-24x+27=10\)
\(\Leftrightarrow-24x=-27\)
\(\Leftrightarrow x=\dfrac{27}{24}\)
Vậy \(x=\dfrac{27}{24}\)
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
a) \(\left(x-3\right)^2+\left(4-x\right)\left(x+4\right)=10\)
\(\Leftrightarrow\left(x^2-2\cdot x\cdot3+3^2\right)+\left(4-x\right)\left(4+x\right)=10\)
\(\Leftrightarrow x^2-6x+9+\left(4^2-x^2\right)-10=0\)
\(\Leftrightarrow x^2-6x-1+16-x^2=0\)
\(\Leftrightarrow-6x+15=0\)
\(\Leftrightarrow6x=15\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
b) \(x^2-2x=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x^2-3^2\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(x+3\right)^2-\left(x-3\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2\left[\left(x+3\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)^2=0\\\left(x+3\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\left(x+3\right)^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=1\\x+3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)
\(a,\dfrac{3}{8}=\dfrac{6}{x}\\ \Rightarrow x=6:\dfrac{3}{8}\\ \Rightarrow x=16\\ b,\dfrac{1}{9}=\dfrac{x}{27}\\ \Rightarrow x=\dfrac{1}{9}.27\\ \Rightarrow x=3\\ c,\dfrac{4}{x}=\dfrac{8}{6}\\ \Rightarrow x=4:\dfrac{4}{3}\\ \Rightarrow x=3\\ d,\dfrac{3}{x-5}=\dfrac{-4}{x+2}\\ \Rightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Rightarrow3x+6=-4x+20\\ \Rightarrow3x+6+4x-20=0\\ \Rightarrow7x-14=0\\ \Rightarrow7x=14\\ \Rightarrow x=2\)
a: =>6/x=3/8
hay x=16
b: =>x/27=1/9
nên x=3
c: =>4/x=4/3
nên x=3
d: =>3/x-5=-4/x+2
=>3x+2=-4x+20
=>7x=18
hay x=18/7
`#040911`
`a)`
\(\left(2x-1\right)^2-\left(2x+5\right)\left(2x+1\right)=10\)
\(\Leftrightarrow 4x^2 - 4x + 1 - (4x^2 + 12x + 5) = 10 \\ \Leftrightarrow 4x^2 - 4x + 1 - 4x^2 - 12x - 5 = 10 \\ \Leftrightarrow (4x^2 - 4x^2) - (4x + 12x) + (1 - 5) = 10 \\ \Leftrightarrow -16x - 4 = 10 \Leftrightarrow -16x = 10 + 4 \\ \Leftrightarrow -16x = 14 \\ \Leftrightarrow x = \dfrac{-7}{8}\)
Vậy, `x = -7/8`
`b)`
`9^2(x - 1) + 25(1 - x) = 0`
`<=> 9^2(x - 1) - 25(x - 1) = 0`
`<=> (x - 1)(9^2 - 5^2) = 0`
`<=>`\(\left[{}\begin{matrix}x-1=0\\9^2-5^2=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=1\\56=0\left(\text{vô lý}\right)\end{matrix}\right.\)
Vậy, `x = 1`
`c)`
`x^2+3x - 4 = 0`
`<=> x^2 + 4x - x - 4 = 0`
`<=> (x^2 - x) + (4x - 4) = 0`
`<=> x(x - 1) + 4(x - 1) = 0`
`<=> (x + 4)(x - 1) = 0`
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\\ \text{Vậy, }x\in\left\{-4;1\right\}\)
a: =>4x^2-4x+1-(4x^2+2x+10x+5)=10
=>4x^2-4x+1-10-4x^2-12x-5=0
=>-16x-4=0
=>x=-1/4
b: =>(x-1)(9^2-25)=0
=>x-1=0
=>x=1
c: =>x^2+4x-x-4=0
=>(x+4)(x-1)=0
=>x=1 hoặc x=-4
a: =>4x^2-24x+36-4x^2+4x-1<10
=>-20x<10-35=-25
=>x>=5/4
b: =>x(x^2-25)-x^3-8<=3
=>x^3-25x-x^3-8<=3
=>-25x<=11
=>x>=-11/25