1) \(\left(x-1\right)^2=25\)
2) \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
3) \(\left(x+20\right)^{100}+|y+4|=0\)
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27 tháng 11 2017
a)
\(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=-\dfrac{1}{4}-y\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}-\dfrac{1}{3}+x=-\dfrac{1}{4}-y\\\dfrac{1}{2}-\dfrac{1}{3}+x=\dfrac{1}{4}+y\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x+y=-\dfrac{5}{12}\\x-y=\dfrac{1}{12}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{6}\\y=-\dfrac{1}{4}\end{matrix}\right.\)
b)\(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
ta thấy : \(\left|x-y\right|\ge0\\ \left|y+\dfrac{9}{25}\right|\ge0\)\(\Rightarrow\left|x-y\right|+\left|y+\dfrac{9}{25}\right|\ge0\)
đẳng thửc xảy ra khi : \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow x=y=-\dfrac{9}{25}\)
vậy \(\left(x;y\right)=\left(-\dfrac{9}{25};-\dfrac{9}{25}\right)\)
27 tháng 11 2017
c) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\)
ta thấy \(\left(\dfrac{1}{2}x-5\right)^{20}\:và\:\left(y^2-\dfrac{1}{4}\right)^{10}\) là các lũy thừa có số mũ chẵn
\(\Rightarrow\:\)\(\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\ \left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)
đẳng thức xảy ra khi \(\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=-\dfrac{1}{2}\\y=\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy cặp số x,y cần tìm là \(\left(10;\dfrac{1}{2}\right)\:hoặc\:\left(10;-\dfrac{1}{2}\right)\)
d)
\(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\\ \Leftrightarrow x\left(x^2-\dfrac{5}{4}\right)=x\left(vì\:x\ge0\right)\\ \Leftrightarrow x\left(x^2-\dfrac{9}{4}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{9}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
vậy x cần tìm là \(-\dfrac{3}{2};0;\dfrac{3}{2}\)
e)\(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
ta thấy: \(x^2\ge0;\left(y-\dfrac{1}{10}\right)^4\ge0\)
\(\Rightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)
đẳng thức xảy ra khi: \(\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
vậy cặp số cần tìm là \(0;\dfrac{1}{10}\)
29 tháng 4 2021
1) Ta có: \(\left(3-x^2\right)+6-2x=0\)
\(\Leftrightarrow3-x^2+6-2x=0\)
\(\Leftrightarrow-x^2-2x+9=0\)
\(\Leftrightarrow x^2+2x-9=0\)
\(\Leftrightarrow x^2+2x+1=10\)
\(\Leftrightarrow\left(x+1\right)^2=10\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)
29 tháng 4 2021
2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)
\(\Leftrightarrow10x-5+7=8-4x+2\)
\(\Leftrightarrow10x+4x=8+2+5-7\)
\(\Leftrightarrow14x=8\)
\(\Leftrightarrow x=\dfrac{4}{7}\)
Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)
25 tháng 7 2021
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
Y
11 tháng 1 2023
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
2
11 tháng 1 2023
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
14 tháng 12 2021
\(\Leftrightarrow\left\{{}\begin{matrix}6x-6-2y+4=0\\4x+4-3y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=1\\4x-3y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)
VM
24 tháng 12 2019
a) \(2^{x-1}=16\)
\(\Rightarrow2^{x-1}=2^4\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=4+1\)
\(\Rightarrow x=5\)
Vậy \(x=5.\)
c) \(\left(x+20\right)^{100}+\left|y+4\right|=0\)
Ta có:
\(\left\{{}\begin{matrix}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\forall x,y.\)
\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\) \(\forall x,y\)
\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x+20=0\\y+4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0-20\\y=0-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-20\\y=-4\end{matrix}\right.\)
Vậy \(\left(x;y\right)\in\left\{-20;-4\right\}.\)
Chúc bạn học tốt!
AN
24 tháng 12 2019
b)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}.\left[1-\left(x-1\right)^4\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^4=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x-1=1\Rightarrow x=2\end{matrix}\right.\)
Vậy ...
Làm cho hết chớ .-.
NH
7 tháng 12 2019
a, \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)\(\Rightarrow x=\frac{5}{6}\)
b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)
\(\Rightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^4=1\end{cases}}\)
Giải: \(\left(x-1\right)^4=1\)\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)
c, Vì \(\left(x+20\right)^{100}\ge0\)\(\forall x\inℝ\); \(\left|y+4\right|\ge0\)\(\forall y\inℝ\)
\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)\(\forall x,y\inℝ\)
Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+20=0\\y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)
d, \(2^{x-1}=16\)\(\Rightarrow2^{x-1}=2^4\)=> x - 1 = 4 => x = 5
1/ \(\left(x-1\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{25}\\x+1=-\sqrt{25}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=5\\x+1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
Vậy...
2/ \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)
\(\Leftrightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+2}=0\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^{x+4}-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^{x+2}=0\\\left(x-1\right)^{x+4}-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)
Vậy...
3/ Với mọi x, y ta có :
\(\left\{{}\begin{matrix}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)
Mà \(\left(x+20\right)^{100}+\left|y+4\right|=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+20=0\\y+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-20\\y=-4\end{matrix}\right.\)
Vậy..
1) (x - 1)2 = 25
(x - 1)2 = 52
=> x - 1 = 5
x = 5 + 1
x = 6