3.Tìm x,biết:
a) (x-2)(x+2)-7=(x-1)2
b) (x-3)2-(x-3)(x+3)=0
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a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)
\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)
hay \(x=-\dfrac{1}{4}\)
c) Ta có: \(8x^3-50x=0\)
\(\Leftrightarrow2x\left(4x^2-25\right)=0\)
\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)
a) \(\left(x-4\right)^2-\left(x-4\right)=0\)
\(\left(x-4\right)\left(x-4-1\right)=0\)
\(\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
b) \(5x^2\left(x-7\right)+7\left(x-7\right)=0\)
\(\left(x-7\right)\left(5x^2+7\right)=0\)
\(\left[{}\begin{matrix}x-7=0\\5x^2+7=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=7\\x^2=\dfrac{-7}{5}\end{matrix}\right.\)
\(x=7\)
c) \(x^2\left(x-3\right)-\left(x-3\right)=0\)
\(\left(x-3\right)\left(x^2-1\right)=0\)
\(\left[{}\begin{matrix}x=3\\x=\pm1\end{matrix}\right.\)
a) (x - 4)^2=(x - 4)
(x - 4) (x -4)=(x -4 )
(x - 4) (x - 4)-(x - 4)=0
(x-4) (x-4-1)=0
(x-4) (x-5)=0
TH1:x-4=0 TH2:x-5=0
x=4 x=5
a) 3/35 - (3/5 + x) = 2/7
=> 3/5 + x= 3/35- 2/7
=> 3/5 +x = -1/5
=> x = -1/5 -3/5
=> x = -4/5
b) 3/7 +1/7 : x = 3/14
=> 1/7 : x= 3/14 -3/7
=> 1/7 : x = -3/14
=> x = 1/7 : -3/14
=> x = -2/3
c) (5x-1).(2x-1/3)=0
=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)
Học tốt :D
a)x=-4/5
b)x=-2/3
c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy.........
mik lười mong bn thông cảm
a) \(x\left(x-6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\left(-7-x\right)\left(-x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-7\\x=-5\end{matrix}\right.\)
c) \(\left(x+3\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=7\end{matrix}\right.\)
d) \(\left(x-3\right)\left(x^2+12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\text{(vô lý)}\end{matrix}\right.\)
\(\Rightarrow x=3\)
e) \(\left(x+1\right)\left(2-x\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x+1\ge0\\2-x\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x+1\le0\\2-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\ge-1\\x\le2\end{matrix}\right.\\\left[{}\begin{matrix}x\le-1\\x\ge2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le2\\x\in\varnothing\end{matrix}\right.\)
\(\Rightarrow-1\le x\le2\)
f) \(\left(x-3\right)\left(x-5\right)\le0\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-3\le0\\x-5\ge0\end{matrix}\right.\\\left[{}\begin{matrix}x-3\ge0\\x-5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le3\\x\ge5\end{matrix}\right.\\\left[{}\begin{matrix}x\ge3\\x\le5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow3\le x\le5\)
a) =>\(\left[{}\begin{matrix}x=0\\x-6=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b => \(\left[{}\begin{matrix}-7-x=0\\-x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)
d) => \(\left[{}\begin{matrix}x-3=0\\x^2+12=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-12\end{matrix}\right.\)(vô lí) => x=3
\(a,=x^2-4-x^2-2x-1=-2x-5\\ b,=8x^3-1-8x^3-1=-2\\ 3,\\ a,\Rightarrow x^3+8-x^3+2x=15\\ \Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\\ b,\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\\ \Rightarrow7x=14\Rightarrow x=2\)
Bài 2:
a) \(=x^2-4-x^2-2x-1=-2x-5\)
b) \(=8x^3-1-8x^3-1=-2\)
Bài 3:
a) \(\Rightarrow x^3+8-x^3+2x=15\)
\(\Rightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)
b) \(\Rightarrow x^3-3x^2+3x-1-x^3+3x^2+4x=13\)
\(\Rightarrow7x=14\Rightarrow x=2\)
(2-x)^3+(2+x)^3-12x(x+1)=0
=>\(8-12x+6x^2-x^3+8+12x+6x^2+x^3-12x\left(x+1\right)=0\)
=>\(12x^2+16-12x^2-12x=0\)
=>16-12x=0
=>4-3x=0
=>x=4/3
a: \(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}\left(x+1\right)\)
=>\(-\dfrac{3}{2}x+\dfrac{1}{4}=\dfrac{1}{2}x+\dfrac{1}{2}\)
=>\(-\dfrac{3}{2}x-\dfrac{1}{2}x=\dfrac{1}{2}-\dfrac{1}{4}\)
=>\(-2x=\dfrac{1}{4}\)
=>\(2x=-\dfrac{1}{4}\)
=>\(x=-\dfrac{1}{4}:2=-\dfrac{1}{8}\)
b: ĐKXĐ: x>=0
\(\left(6-3\sqrt{x}\right)\left(\left|x\right|-7\right)=0\)
=>\(\left\{{}\begin{matrix}6-3\sqrt{x}=0\\\left|x\right|-7=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}3\sqrt{x}=6\\\left|x\right|=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-7\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
a) (x - 3)2 - 5.(x - 2) + 5 = 0.
<=> x^2 - 6x + 9 - 5x + 10 + 5 = 0
<=> x^2 - 11x + 24 = 0
<=> (x-3)(x-8)=0
<=> x = 3 hoặc x = 8
a: x-2b+c=1-2+3=2
b: x-2b+c=5-2*0+2=7
c: x-2b+c=3-2*2-4=-5
b) (x-3)2-(x-3)(x+3)=0
<=>(x-3)[(x-3)-(x+3)]=0
<=>(x-3)[x-3-x-3]=0
<=>(x-3)*(-6)=0
<=>x=3
a) (x-2)(x+2)-7=(x-1)2
<=>x2-4-7=x2-2x+1
<=>x2-11-x2+2x-1=0
<=>(x2-x2)+2x-11-1=0
<=>2x-12=0
<=>2x=12
<=>x=6