Tìm x :
( 36x2 - 25 ) - ( 6x + 5 ) ( x + 1) = 0
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7 tháng 11 2021
\(a,\Leftrightarrow6x-9+4-2x=-3\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(x-2021\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2021\\x=6\end{matrix}\right.\\ c,\Leftrightarrow\left(2x-3-6x\right)\left(2x-3+6x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-3-4x=0\\8x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{8}\end{matrix}\right.\)
C
1
13 tháng 6 2021
`a)(x-6)^2-(x+6)^2=12`
`<=>(x-6-x-6)(x-6+x+6)=12`
`<=>-12.2x=12`
`<=>2x=-1`
`<=>x=-1/2`
Vậy `x=-1/2`
`b)36x^2-12x+1=81`
`<=>(6x-1)^2=81`
`<=>(6x-1-9)(6x-1+9)=0`
`<=>(6x-10)(6x+8)=0`
`<=>(3x-5)(3x+4)=0`
`<=>` \(\left[ \begin{array}{l}x=\dfrac53\\x=-\dfrac43\end{array} \right.\)
`c)x^2-4x-12=0`
`<=>x^2-6x+2x-12=0`
`<=>x(x-6)+2(x-6)=0`
`<=>(x-6)(x+2)=0`
`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\)
`d)x^2-5x-6=0`
`<=>x^2-6x+x-6=0`
`<=>x(x-6)+x-6=0`
`<=>(x-6)(x+1)=0`
`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\)
ND
Nguyễn Đức Trí
VIP
23 tháng 8 2023
a) \(A\left(x\right)=x^2-10x+25\)
\(\Rightarrow A\left(x\right)=\left(x-5\right)^2\)
\(\Rightarrow\left\{{}\begin{matrix}A\left(0\right)=\left(0-5\right)^2=25\\A\left(-1\right)=\left(-1-5\right)^2=36\end{matrix}\right.\)
b) \(A\left(x\right)+B\left(x\right)=6x^2-5x+25\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-A\left(x\right)\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-\left(x^2-10x+25\right)\)
\(\Rightarrow B\left(x\right)=6x^2-5x+25-x^2+10x-25\)
\(\Rightarrow B\left(x\right)=5x^2+5x\)
\(\Rightarrow B\left(x\right)=5x\left(x+1\right)\)
c) \(A\left(x\right)=\left(x-5\right)C\left(x\right)\)
\(\Rightarrow C\left(x\right)=\dfrac{\left(x-5\right)^2}{x-5}=x-5\left(x\ne5\right)\)
d) Nghiệm của B(x)
\(\Leftrightarrow B=0\)
\(\Leftrightarrow5x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) là nghiệm của B(x)
DT
4
NN
8 tháng 12 2019
\(2x\left(x^2-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
\(2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\left(2x+1\right)\left(3x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)
NN
8 tháng 12 2019
\(9\left(3x-2\right)-x\left(2-3x\right)=0\)
\(9\left(3x-2\right)+x\left(3x-2\right)=0\)
\(\left(9+x\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)
\(\left(2x-1\right)^2=25\)
\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
AH
4 tháng 10 2016
A , 6x - 5 = 613
6x = 618
x = 103
b , x - 48 : 16 = 25
x - 3 = 25
x = 28
c , ( x - 8 ) . ( x - 7 ) = 0
=> x - 8 = 0 hoặc x - 7 = 0
x = 8 x = 7
Vậy x = 8 hoặc x = 7
20 tháng 8 2021
1, \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow x=0;x=\pm5\)
2, \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+9\right)=0\Leftrightarrow x=-9;x=1\)
3, \(6x\left(x-2\right)=x-2\Leftrightarrow\left(6x-1\right)\left(x-2\right)=0\Leftrightarrow x=\frac{1}{6};x=2\)
4, \(7\left(x-2020\right)^2-x+2020=0\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)
\(\Leftrightarrow\left(x-2020\right)\left[7\left(x-2020\right)-1\right]=0\Leftrightarrow x=2020;x=\frac{14141}{7}\)
5, \(x^2-10x=-25\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
6, \(x^2-2x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow x=-1;x=3\)
20 tháng 8 2021
\(1,\)
\(2x^3-50x=0\)
\(\Leftrightarrow2x\left(x^2-25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
\(2,\)
\(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\)
\(\Leftrightarrow x^2-x+9x-9=0\)
\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)
\(3,\)
\(6x\left(x-2\right)=x-2\)
\(\Leftrightarrow6x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{6}\end{cases}}\)
\(4,\)
\(7\left(x-2020\right)^2-x+2020=0\)
\(\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)
\(\Leftrightarrow\left(x-2020\right)[7\left(x-2020\right)-1]=0\)
\(\Leftrightarrow\left(x-2020\right)[7x-14141]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\7x=14141\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{14141}{7}\end{cases}}\)
\(5,\)
\(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
\(6,\)
\(x^2-2x-3=0\)
\(\Leftrightarrow x^2-3x+x-3=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
PT
1
CM
20 tháng 3 2017
Ta có:
Vậy điều kiện để phân thức xác định là x ≠ 0 và x ≠ 1.
\(\left(36x^2-25\right)-\left(6x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(6x-5\right)-\left(6x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(6x-5-x-1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(5x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{6}\\x=\frac{6}{5}\end{cases}}\)
\(\left(36x^2-25\right)-\left(6x+5\right)\left(x+1\right)=0\Leftrightarrow\left(6x-5\right)\left(6x+5\right)-\left(6x+5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(6x+5\right)\left(5x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{-5}{6}\\x=\frac{6}{5}\end{cases}}\)