2.Tìm x, biết
a,3.x+7=25
b,(5.x+4)-1=23
c,96-3x(x+1)=42
d,2x=16
e,x2=16
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19 tháng 2 2022
b: \(\Leftrightarrow9x^2+12x+4-18x+12=9x^2\)
=>-6x+16=0
=>-6x=-16
hay x=8/3(nhận)
c: \(\Leftrightarrow\dfrac{x+1+x-1}{\left(x-1\right)\left(x+1\right)}=\dfrac{2}{x+2}\)
\(\Leftrightarrow2x\left(x+2\right)=2\left(x^2-1\right)\)
\(\Leftrightarrow2x^2+4x-2x^2+2=0\)
=>4x+2=0
hay x=-1/2(nhận)
CM
24 tháng 9 2018
a) Rút gọn được VT = 9x + 7. Từ đó tìm được x = 1.
b) Rút gọn được VT = 2x + 8. Từ đó tìm được x = 7 2 .
27 tháng 8 2021
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
27 tháng 8 2021
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
9 tháng 11 2021
Bài 3:
\(\Leftrightarrow x^3+64-x^3+25x=264\)
hay x=8
9 tháng 11 2021
\(1,C=6x^2+23x-55-6x^2-23x-21=-76\\ 2,=\left(2x^4-x^2+2x^3-x-6x^2+6-3\right):\left(2x^2-1\right)\\ =\left[\left(2x^2-1\right)\left(x^2+x-6\right)-3\right]:\left(2x^2-1\right)\\ =x^2+x-6\left(dư.-3\right)\\ 3,\Leftrightarrow x^3+64-x^3+25x=264\\ \Leftrightarrow25x=200\Leftrightarrow x=8\)
19 tháng 7 2021
a) Ta có: \(x^2-2x+1=25\)
\(\Leftrightarrow\left(x-1\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b) Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
c) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
d) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)
\(\Leftrightarrow24x=-10\)
hay \(x=-\dfrac{5}{12}\)
12 tháng 8 2021
a,\(< =>\left(x-1\right)^2-5^2=0< =>\left(x-1-5\right)\left(x-1+5\right)=0\)
\(< =>\left(x-6\right)\left(x+4\right)=0=>\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
b,\(< =>25x^2+10x+1-25x^2+9-30=0\)
\(< =>10x-20=0< =>10\left(x-2\right)=0< =>x=2\)
c,\(< =>x^3-1-x\left(x^2-4\right)-5=0\)
\(< =>x^3-1-x^2+4x-5=0< =>4x-6=0< =>x=\dfrac{6}{4}\)\(d,< =>\left(x-2\right)^3-x^3+3^3+6x^2+12x+6-15=0\)
\(< =>x^3-6x^2+12x-x^3+6x^2+12x+10=0\)
\(< =>24x+10=0< =>x=-\dfrac{5}{12}\)
12 tháng 8 2021
a: Ta có: \(x^2-2x+1=25\)
\(\Leftrightarrow\left(x-4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=6\end{matrix}\right.\)
b: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)
\(\Leftrightarrow10x=20\)
hay x=2
c: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x=5\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)
10 tháng 12 2021
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
CM
11 tháng 9 2019
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
a) 3.x+7=25
3.x = 25 -7
. 3.x = 18
x. = 18 :3
. x. 6
a. 3.x+7=25
<=> 3x = 18
<=> x = 6
b. (5.x+4)-1=23
<=> 5x + 4 = 24
<=> 5x = 20
<=> x = 4
c. 96-3(x+1)=42
<=> 3(x + 1) = 54
<=> x + 1 = 18
<=> x = 17
d. 2x=16
<=> 2x = 24
<=> x = 4
e. x2=16
<=> \(\orbr{\begin{cases}x^2=4^2\\x^2=\left(-4\right)^2\end{cases}}\)
<=> x = \(\pm\)4