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20 tháng 12 2018
h) \(\left(x-1\right)^2=25\)
Mà:\(5^2=\left(-5\right)^2=25\)
TH1:\(x-1=5\)
\(x=5+1\)
\(x=6\)
TH2:\(x-1=-5\)
\(x=-5+1\)
\(x=-4\)
Vậy:\(x=6\)hoặc \(x=-4\)
20 tháng 12 2018
i)\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
Mà:\(\left(\frac{2}{5}\right)^2=\left(\frac{-2}{5}\right)^2=\frac{4}{25}\)
TH1:\(x+\frac{1}{2}=\frac{2}{5}\) TH2:\(x+\frac{1}{2}=\frac{-2}{5}\)
\(x=\frac{2}{5}-\frac{1}{2}\) \(x=\frac{-2}{5}-\frac{1}{2}\)
\(x=\frac{-1}{10}\) \(x=\frac{-9}{10}\)
Vậy:\(x=\frac{-1}{10}\)hoặc\(x=\frac{-9}{10}\)
19 tháng 9 2016
1) Các cách viết số 25 dưới dãng lũy thừa là: 251; 52; (-5)2
2) a) \(\left(x-\frac{1}{2}\right)^2=0\)
=> \(x-\frac{1}{2}=0\)
=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
b) (x - 2)2 = 1
=> \(\left[\begin{array}{nghiempt}x-2=1\\x-2=-1\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=3\\x=1\end{array}\right.\)
Vậy \(x\in\left\{3;1\right\}\)
c) (2x - 1)3 = -8
=> (2x - 1)3 = (-2)3
=> 2x - 1 = -2
=> 2x = -2 + 1
=> 2x = -1
=> \(x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=16\)
=> \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{array}\right.\)
Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}\)
P
14 tháng 11 2018
1) Các cách viết số 25 dưới dãng lũy thừa là: 251; 52; (-5)2
2) a) (x−12)2=0(x−12)2=0
=> x−12=0x−12=0
=> x=12x=12
Vậy x=12x=12
b) (x - 2)2 = 1
=> [x−2=1x−2=−1[x−2=1x−2=−1=> [x=3x=1[x=3x=1
Vậy x∈{3;1}x∈{3;1}
c) (2x - 1)3 = -8
=> (2x - 1)3 = (-2)3
=> 2x - 1 = -2
=> 2x = -2 + 1
=> 2x = -1
=> x=−12x=−12
Vậy x=−12x=−12
d) (x+12)2=16(x+12)2=16
=> [x+12=14x+12=−14[x+12=14x+12=−14=> [x=−14x=−34[x=−14x=−34
Vậy x∈{−14;−34}
PA
8 tháng 7 2016
a.
\(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\)
\(6x^2+21x-2x-7-6x^2+5x-6x+5=16\)
\(\left(6x^2-6x^2\right)+\left(21x-2x+5x-6x\right)-\left(7-5\right)=16\)
\(18x-2=16\)
\(18x=16+2\)
\(18x=18\)
\(x=\frac{18}{18}\)
\(x=1\)
b.
\(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(10x^2+9x-10x^2-15x+2x+3=8\)
\(\left(10x^2-10x^2\right)-\left(15x-9x-2x\right)+3=8\)
\(-4x=8-3\)
\(-4x=5\)
\(x=-\frac{5}{4}\)
c.
\(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)
\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(\left(15x^2-15x^2\right)+\left(25x+21x-10x+6x\right)-\left(35+4+2\right)=0\)
\(42x=41\)
\(x=\frac{41}{42}\)
TT
0
28 tháng 12 2020
Bài 1 :
\(\frac{x-1}{x-5}=\frac{6}{7}\Leftrightarrow7x-7=6x-30\)
\(\Leftrightarrow x=-23\)
\(\frac{x-2}{x-1}=\frac{x+4}{x+7}\)ĐK : \(x\ne1;-7\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=\left(x+4\right)\left(x-1\right)\)
\(\Leftrightarrow x^2+5x-14=x^2+3x-4\)
\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)
NN
3 tháng 5 2020
1. \(\left(3x-5\right)^{2010}+\left(y-1\right)^{2012}+\left(x-z\right)^{2014}=0\)
Vì \(\left(3x-5\right)^{2010}\ge0\forall x\); \(\left(y-1\right)^{2012}\ge0\forall y\); \(\left(x-z\right)^{2014}\ge0\forall x,z\)
\(\Rightarrow\left(3x-5\right)^{2010}+\left(y-1\right)^{2012}+\left(x-z\right)^{2014}\ge0\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}3x-5=0\\y-1=0\\x-z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}3x=5\\y=1\\x=z\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{3}\\y=1\\z=\frac{5}{3}\end{cases}}\)
Vậy \(x=z=\frac{5}{3}\)và \(y=1\)
TA
2
VM
7 tháng 10 2019
a) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow x+\frac{1}{2}=\pm\frac{1}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{4}-\frac{1}{2}\\x=\left(-\frac{1}{4}\right)-\frac{1}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{4};-\frac{3}{4}\right\}.\)
b) \(\left(3x+1\right)^3=-27\)
\(\Rightarrow\left(3x+1\right)^3=\left(-3\right)^3\)
\(\Rightarrow3x+1=-3\)
\(\Rightarrow3x=\left(-3\right)-1\)
\(\Rightarrow3x=-4\)
\(\Rightarrow x=\left(-4\right):3\)
\(\Rightarrow x=-\frac{4}{3}\)
Vậy \(x=-\frac{4}{3}.\)
Mấy câu sau làm tương tự nhé.
Chúc bạn học tốt!
NB
7 tháng 10 2019
c)\(\left(3x-2\right)^2=36\\ \Leftrightarrow\left(3x-2\right)^2=\left(\pm6\right)^2\\ \Rightarrow\left\{{}\begin{matrix}3x-2=6\\3x-2=-6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{8}{3}\\x=-\frac{4}{3}\end{matrix}\right.\)
d)\(\left(\frac{2}{5}-3x\right)^2=\frac{9}{25}\\ \Leftrightarrow\left(\frac{2}{5}-3x\right)^2=\left(\pm\frac{3}{5}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}\frac{2}{5}-3x=\frac{3}{5}\\\frac{2}{5}-3x=-\frac{3}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\frac{1}{15}\\x=\frac{1}{3}\end{matrix}\right.\)
26 tháng 9 2015
a) \(-0,6^0+\frac{1}{2}.2-3x=-\frac{1}{4}\)
\(\Leftrightarrow-1+1-3x=-\frac{1}{4}\Leftrightarrow-3x=-\frac{1}{4}\Leftrightarrow3x=\frac{1}{4}\Leftrightarrow x=\frac{1}{4}:3=\frac{1}{12}\)
b)\(2^{x-2}+22=3.2^x\Leftrightarrow3.2^x-2^{x-2}=22\Leftrightarrow2^{x-2}\left(3.2^2-1\right)=22\)
\(\Leftrightarrow2^{x-2}.11=22\Leftrightarrow2^{x-2}=2\Leftrightarrow x-2=1\Leftrightarrow x=3\)
c) \(\left(x-1\right)^2=\sqrt{\left(-\frac{9}{16}\right)^2}\Leftrightarrow\left(x-1\right)^2=\frac{9}{16}\Leftrightarrow\left(x-1\right)^2=\left(\frac{3}{4}\right)^2\)
TH1: x - 1 = 3/4 => x = 3/4 + 1 => x = 7/4
Th2: x - 1 = - 3/4 => x = -3/4 +1 => x = 1/4
d) \(\Leftrightarrow\sqrt{x^2+2}=12-5=7\Leftrightarrow x^2+2=7^2\Leftrightarrow x^2=49-2\Leftrightarrow x^2=47\)
\(x=\sqrt{47};x=-\sqrt{47}\)
tích đúng mình giải cho
Ta có: \(\left(x+\frac{1}{2}\right)^2-\frac{1}{16}=0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
Mà \(\frac{1}{16}=\left(\frac{1}{4}\right)^2\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\Rightarrow x=\frac{-1}{4}\)
Vậy ....
\(\left(3x+\frac{1}{2}\right)^2+\frac{25}{16}=0\)
\(\Rightarrow\left(3x+\frac{1}{2}\right)^2=\frac{-25}{16}\)
Vì \(\left(3x+\frac{1}{2}\right)^2\ge0\left(\forall x\in Z\right)\)
Nên x thuộc rỗng (không có giá trị của x)