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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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)
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a. 5 - 3(x + 4) = -1
⇔ 5 - 3x - 12 = -1
⇔ 3x = -1 - 5 + 12
⇔ 3x = 6
⇔ x = 2
\(d,2x^2-3=5\)
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x=\pm2\)
\(e,x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Hai bài bị trùng nhau nên các bạn nhìn ảnh hay văn bản đều như nhau ạ
c: =>x+2>0
hay x>-2
d: =>-4<=x<=3
e: =>\(x\in\varnothing\)
f: \(\Leftrightarrow\left[{}\begin{matrix}x>4\\x< -6\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. $(x^2-9)(5x+15)=0$
$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$
$\Rightarrow x^2=9=3^2=(-3)^2$
$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$
$\Rightarrow x=-3$
b.
$x^2-8x=0$
$\Rightarrow x(x-8)=0$
$\Rightarrow x=0$ hoặc $x-8=0$
$\Rightarrow x=0$ hoặc $x=8$
c.
$5+12(x-1)^2=53$
$12(x-1)^2=53-5=48$
$(x-1)^2=48:12=4=2^2=(-2)^2$
$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$
d.
$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$
$\Rightarrow x=11$ hoặc $x=-1$
e.
$(3x-5)^3=64=4^3$
$\Rightarrow 3x-5=4$
$\Rightarrow 3x=9$
$\Rightarrow x=3$
f.
$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$
$2^{4x}(1+8)=144$
$2^{4x}.9=144$
$2^{4x}=144:9=16=2^4$
$\Rightarrow 4x=4\Rightarrow x=1$
![](https://rs.olm.vn/images/avt/0.png?1311)
`x^2=3`
`=>x=\sqrt{3}\or\x=-\sqrt{3}`
`x^2=36`
`<=>x^2=(+-6)^2`
`<=>x=+-6`
`x^2=25`
`<=>x^2=(+-5)^2`
`<=>x=+-5`
`2x^2+(-20)=55`
`<=>2x^2-20=55`
`<=>2x^2=75`
`<=>x^2=75/2`
`<=>x=+-\sqrt{75/2}`
`2(x-1)^2+5^0=9`
`<=>2(x-1)^2+1=9`
`<=>2(x-1)^2=8`
`<=>(x-1)^2=4`
`<=>x-1=2\or\x-1=-2`
`<=>x=3\or\x=-1`
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1).( 27,56 x 35 ) + ( 27,56 x 67 ) - ( 27,56 x 2)
= (964 + 1846,52) - 55,12
=2810,52 - 55,12
= 2755,4
2).( 4x 35 ) x ( 25 x 5 ) x 2
= ( 140 x 125 ) x2
= 17500 x 2
=35000
4). 3/10
5). 1188
6). 61/6
![](https://rs.olm.vn/images/avt/0.png?1311)
X2=3 x2=25
=> X=\(\pm\sqrt{3}\) => x=5
X2=36
=> x=6
2.(x-1)2+50= 9
2.(x-1)2+1= 9
2.(x-1)2= 8
(x-1)2 = 8/2
(x-1)2 = 4
(x-1)2 = (2)2
x-1=(\(\pm\)2)
TH1: x-1= 2 TH2: x-1=-2
x=2+1 x =(-2)+1
x= 3 x = -1
Vậy x\(\in\)\(\left\{3;1\right\}\)
\(\left(x-1\right)x\left(x+3\right)=0\)
khi ít nhất:1 trong 3 thừa số trên là: 0
dễ thấy chỉ có thể có 1 số bằng 0 và 2 số còn lại khác 0
+) x-1=0=>x=1
+)x+3=0=>x=-3
TH còn lại là: x=0
Vậy: x E {0;-3;1}
b)c)d) tt
\(a,\left(x-1\right).\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)
\(b,\left(x^2-1\right).\left(x^2-81\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-9\right)\left(x+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{x-1=0}{x+1=0}\\\frac{x-9=0}{x+9=0}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{x=1}{x=-1}\\\frac{x=9}{x=-9}\end{cases}}\)
\(c,\left(x^2+1\right)\left(x^2-125\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-125=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2=1\left(voly\right)\\x^2=125\end{cases}}\)
\(\Leftrightarrow x=5\sqrt{5}\)
\(d,\left(x-5\right)^2=36\Leftrightarrow x-5=\pm6\Leftrightarrow x=-1;11\)