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bài 1)
a) \(\dfrac{11}{13}-\left(\dfrac{5}{42}-x\right)=-\left(\dfrac{15}{28}-\dfrac{11}{15}\right)
\)
\(\left(\dfrac{5}{42}-x\right)=\dfrac{11}{13}+\dfrac{15}{28}-\dfrac{11}{15}\)
\(x=\dfrac{5}{42}-\dfrac{3541}{5460}=-\dfrac{413}{780}\)
b) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|2,15\right|\)
\(\left|x+\dfrac{4}{15}\right|=-\left|2,15\right|+\left|3,75\right|=1,6\)
\(\Rightarrow x+\dfrac{4}{15}=1,6\) hoặc \(x+\dfrac{4}{15}=-1,6\)
\(\Rightarrow x=\dfrac{4}{3}\) hoặc \(x=-\dfrac{28}{15}\)
c) \(\dfrac{5}{3}-\left|x-\dfrac{3}{2}\right|=-\dfrac{1}{2}\)
\(\Rightarrow\left|x-\dfrac{3}{2}\right|=\dfrac{5}{3}+\dfrac{1}{2}=\dfrac{13}{6}\)
\(\Rightarrow x-\dfrac{3}{2}=\dfrac{13}{6}\) hoặc \(x-\dfrac{3}{2}=-\dfrac{13}{6}\)
\(\Rightarrow x=\dfrac{11}{3}\) hoặc \(x=-\dfrac{2}{3}\)
d)\(\left(x-\dfrac{2}{3}\right).\left(2x-\dfrac{3}{2}\right)=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\) hoặc \(2x-\dfrac{3}{2}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{3}{4}\end{matrix}\right.\)
3) a) \(\left(x^{^2}-4\right)^{^2}+\left(x+2\right)^{^2}=0\)
Vì \(\left(x^{^2}-4\right)^{^2}\ge0,\left(x+2\right)^{^2}\ge0\) nên :
\(\left\{{}\begin{matrix}x^{^2}-4=0\\x+2=0\end{matrix}\right.\Rightarrow x=\pm2\)
b) \(\left(x-y\right)^{^2}+\left|y+2\right|=0\)
Vì \(\left\{{}\begin{matrix}\left(x-y\right)^{^2}\ge0\\\left|y+2\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-y=0\\y=-2\end{matrix}\right.\Rightarrow x=-2;y=-2\)
c) \(\left|x-y\right|+\left|y+\dfrac{9}{25}\right|=0\)
Vì \(\left\{{}\begin{matrix}\left|x-y\right|\ge0\\\left|y+\dfrac{9}{25}\right|\ge0\end{matrix}\right.\) nên \(\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Rightarrow y=-\dfrac{9}{25};x=-\dfrac{9}{25}\)
d) \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|=\left(-\dfrac{1}{4}\right)-\left|y\right|\)
\(\Rightarrow\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\)
Vì \(\left\{{}\begin{matrix}\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|\ge0\\\left|y\right|\ge0\end{matrix}\right.\) mà \(\left|\dfrac{1}{2}-\dfrac{1}{3}+x\right|+\left|y\right|=-\dfrac{1}{4}\) nên không tồn tại x,y thỏa mãn đề bài .
1) |x|=x+2
=> \(\left[{}\begin{matrix}x=x+2\\x=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2\left(voli\right)\\2x=-2\Rightarrow x=-1\end{matrix}\right.\)
vậy x=-1
c;b tương tự
2) \(\left|x-\dfrac{3}{2}\right|=\left|\dfrac{5}{2}-x\right|\)
=> \(\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{5}{2}-x\\x-\dfrac{3}{2}=x-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=4\Rightarrow x=2\\0=-1\left(voli\right)\end{matrix}\right.\)
vậy x=2
\(A=2x^2-2\ge-2\)
Dấu "=" xảy ra khi: \(x=0\)
\(B=\left|x+\dfrac{1}{3}\right|-\dfrac{1}{6}\ge-\dfrac{1}{6}\)
Dấu "=" xảy ra khi: \(x=-\dfrac{1}{3}\)
\(C=\dfrac{\left|x\right|+2017}{2018}\ge\dfrac{2017}{2018}\)
Dấu "=" xảy ra khi: \(x=0\)
\(D=3-\left(x+1\right)^2\le3\)
Dấu "=" xảy ra khi: \(x=-1\)
\(E-\left|0,1+x\right|-1,9\le-1,9\)
Dấu "=" xảy ra khi: \(x=-0,1\)
\(F=\dfrac{1}{\left|x\right|+2017}\le\dfrac{1}{2017}\)
Dấu "=" xảy ra khi: \(x=0\)
MinA = 29 \(\Leftrightarrow x=0\)
Min B= 625 \(\Leftrightarrow x=\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
a: TH1: x>=0
=>x+x=1/3
=>x=1/6(nhận)
TH2: x<0
Pt sẽ là -x+x=1/3
=>0=1/3(loại)
b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
c: \(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-8}+\dfrac{1}{x-8}-\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{2}{x-20}=\dfrac{-3}{4}\)
\(\Leftrightarrow\dfrac{x-20-2x+2}{\left(x-1\right)\left(x-20\right)}=\dfrac{-3}{4}\)
\(\Leftrightarrow-3\left(x^2-21x+20\right)=4\left(-x-18\right)\)
\(\Leftrightarrow3x^2-63x+60=4x+72\)
=>3x^2-67x-12=0
hay \(x\in\left\{22.51;-0.18\right\}\)
bài 1 :
b) (x-1/2 )2 = 0
<=> x - 1/2 = 0
<=> x = 0+ 1/2
<=> x = 1/2
c) ( x - 2 ) 2 = 1
<=> x -2 = 1
<=> x = 1 +2 = 3
d) ( 2x -1 )3 = -8
<=> ( 2x - 1) 3 = ( -2 ) 3
<=> 2x - 1 = -2
<=> 2x = -2+1 = -1
<=> x = -1/2
Bài 2 :
c) 32x-1=243
<=> 32x-1= 35
<=> 2x-1 = 5
<=> 2x = 6
<=> x = 6:2 = 3
Mk chỉ giải đc như vậy thôi
bạn thông cảm nhé !
2) a) \(\left(x+\dfrac{4}{5}\right)^2=\dfrac{9}{25}\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{3}{5}\\x+\dfrac{4}{5}=-\dfrac{3}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{5}\\x=\dfrac{-7}{5}\end{matrix}\right.\) vậy \(x=\dfrac{-1}{5};x=\dfrac{-7}{5}\)
b) \(\left|x-\dfrac{3}{7}\right|=-2\) vì giá trị đối không âm được nên phương trình này vô nghiệm
c) điều kiện : \(x\ge-7\) \(\sqrt{x+7}-2=4\Leftrightarrow\sqrt{x+7}=4+2=6\)
\(\Leftrightarrow x+7=6^2=36\Leftrightarrow x=36-7=29\) vậy \(x=29\)
d) \(x^2-\dfrac{7}{9}x=0\Leftrightarrow x\left(x-\dfrac{7}{9}\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-\dfrac{7}{9}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=\dfrac{7}{9}\end{matrix}\right.\) vậy \(x=0;x=\dfrac{7}{9}\)
1) tìm GTNN
a) \(B=\left|x-2017\right|+\left|x-20\right|\)
B \(\ge\left|x-2017-x+20\right|=\left|-1997\right|=1997\)
Dấu " = " xảy ra khi và chỉ khi 20 \(\le x\le2017\)
Vậy MinB = 1997 khi 20 \(\le x\le2017\)
b) \(C=\left|x-3\right|+\left|x-5\right|\)
\(C\ge\left|x-3-x+5\right|=\left|2\right|=2\)
Dấu " = " xảy ra khi 3 \(\le x\le5\)
Vậ MinC = 2 khi và chỉ khi 3 \(\le x\le5\)
c) \(C=\left|x^2+4\right|+3\)
Ta thấy \(x^2+4\ge0\) với mọi x
nên \(\left|x^2+4\right|+3=x^2+4+3=x^2+7\)\(\ge\) 7
Dấu " =" xảy ra khi x = 0
MinC = 7 khi và chỉ khi x = 0
| x | - 1 = 2
| x | = 2 + 1
|x| = 3
X = 3 hoặc x = -3
chọn D. x = 3 hoặc x = - 3