Tìm x biết:
a, 60%x - x + 1/3 = 2
b, ( 4x^2 - 25 )( x + 3 ) = 0
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`@` `\text {Ans}`
`\downarrow`
`a,`
`(x - 2)(x - 3) =0`
`<=>`\(\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0+2\\x=0+3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy, `S = {2; 3}`
`b,`
`x^2 - 5x = 0`
`<=> x(x - 5) = 0`
`<=>`\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=0+5\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
Vậy, `S = {0; 5}`
`c,`
`x^2 - 9 = 0`
`<=> x^2 = 0 + 9`
`<=> x^2 = 9`
`<=> x^2 = (+-3)^2`
`<=> x = +-3`
Vậy, `S = {3; -3}`
`d,`
`4x^2 - 25 = 0`
`<=> 4x^2 = 25`
`<=> x^2 = 25/4`
`<=> x^2 = (+-5/2)^2`
`<=> x = +-5/2`
Vậy,` S = {5/2; -5/2}.`
a: =>x-2=0 hoặc x-3=0
=>x=2 hoặc x=3
b: =>x(x-5)=0
=>x=0 hoặc x=5
c: =>(x-3)(x+3)=0
=>x=3 hoặc x=-3
d: =>(2x-5)(2x+5)=0
=>x=5/2 hoặc x=-5/2
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
`2(x-3)-4x=0`
`<=> 2x-6-4x=0`
`<=> -2x-6=0`
`<=>-2x=6`
`<=>x=-3`
__
`x^2-2x+1=25`
`<=>(x-1)^2=25`
`<=> (x-1)^2 = (+- 5)^2`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
a: 2(x-3)-4x=0
=>2x-6-4x=0
=>-2x-6=0
=>2x+6=0
=>2x=-6
=>x=-3
b: \(x^2-2x+1=25\)
=>\(\left(x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
a, 70 - 5 ( x - 3 ) = 45
⇔ 5 . x - 3 = 70 - 45
⇔ 5 x - 3 = 35
⇔ x - 3 = 35 : 5 ⇔ x - 3 = 7
⇔ x = 10
b, 10 + 2 x = 4 5 : 4 3
⇔ 10 + 2 x = 4 2 ⇔ 10 + 2 x = 16
⇔ 2 x = 4 ⇔ x = 2
c, 60 - 3 x - 2 = 51
⇔ 3 x - 2 = 60 - 51
⇔ 3 x - 2 = 9
⇔ x - 2 = 3 ⇔ x = 5
d, 4 x - 20 = 2 5 : 2 3
⇔ 4 x - 20 = 2 2 ⇔ 4 x - 20 = 4
⇔ 4 x = 24 ⇔ x = 6
Bài 1:
a) \(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)
\(114x^2+216x+81=114x^2-480x+400\)
\(144x^2+216x=144x^2-480x+400-81\)
\(114x^2+216=114x^2-480x+319\)
\(696x=319\)
\(\Rightarrow x=\frac{11}{24}\)
b) \(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\left(x-1\right)^2\left(x^2+2\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)
\(\Rightarrow x=1\)
c) \(x^5+x^4+x^3+x^2+x+1=0\)
\(\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow x=-1\)
Bài 2:
a) \(5x^3-7x^2-15x+21=0\)
\(\left(5x-7\right)\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)
\(\Rightarrow x=\frac{7}{5}\)
b) \(\left(x-3\right)^2=4x^2-20x+25\)
\(x^2-6x+9-25=4x^2-20x+25\)
\(x^2-6x+9=4x^2-20x+25-25\)
\(x^2-6x-16=4x^2-20x\)
\(x^2+14x-16=4x^2-4x^2\)
\(-3x^2+14x-16=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)
c) \(\left(x-1\right)^2-5=\left(x+2\right)\left(x-2\right)-x\left(x-1\right)\)
\(x^2-2x=x-4\)
\(x^2-2x=x-4+4\)
\(x^2-2x=x-x\)
\(x^2-3x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
d) \(\left(2x-3\right)^3-\left(2x+3\right)\left(4x^2-1\right)=-24\)
\(-48x^2+56x-24=-24\)
\(-48x^2+56x=-24+24\)
\(-48x^2+56=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{6}\end{cases}}\)
mình ko chắc
cảm ơn nha
khôn :v