tim x biet : 4-|x-1/5|=-1/2
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20 . 2^x + 1 = 10.4^2 + 1
20 . 2^x + 1 = 10 . 16 + 1
20 . 2^x + 1 = 161
20 . 2^x = 161 - 1
20 . 2^x = 160
2^x = 8
2^x = 2^3
=> x = 3
a, \(\left|2x+1\right|=5\Rightarrow2x+1\in\left\{5;-5\right\}\)
+) Nếu :\(2x+1=5\Rightarrow2x=4\Rightarrow x=4\div2=2\)
+) Nếu : \(2x+1=-5\Rightarrow2x=-6\Rightarrow x=-6\div2=-3\)
Vậy \(x\in\left\{2;-3\right\}\)
b, \(\left|x-4\right|=\left|2-x\right|\)
\(\Rightarrow\left[\begin{matrix}x-4=2-x\\x-4=-\left(2-x\right)\end{matrix}\right.\)
+) Nếu : x - 4 = 2 - x
\(\Rightarrow x+x=2+4\Rightarrow2x=6\Rightarrow x=3\)
+) Nếu : x - 4 = - ( 2 - x )
\(\Rightarrow x-4=-2+x\Rightarrow x-x=-2+4\Rightarrow0=2\) ( loại )
Vậy x = 3 thỏa mãn đề bài
c, \(\left|x-5\right|=2-x\Rightarrow\left|x-5\right|+x=2\)
+) Nếu : \(x< 5\Rightarrow x-5< 5-5\Rightarrow x-5< 0\Rightarrow\left|x-5\right|=-x+5\)
Thay vào đề , ta có :
\(-x+5+x=2\Rightarrow-x+x+5=2\Rightarrow5=2\) ( loại )
+) Nếu : \(x\ge5\Rightarrow x-5\ge5-5\Rightarrow x-5\ge0\Rightarrow\left|x-5\right|=x-5\)
Thay vào đề , ta có :
\(\left(x-5\right)-x=2\Rightarrow x-5-x=2\)
\(\Rightarrow x-x-5=2\Rightarrow-5=2\) ( loại )
Vậy \(x\in\varnothing\)
\(2\left(|x-1|+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)
\(TH1:x\ge1\Rightarrow|x-1|=x-1\)
\(\Rightarrow2\left(x-1+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)
\(\Rightarrow2\left(2x-\frac{9}{5}\right)=2x-\frac{2}{5}\Rightarrow4x-\frac{18}{5}=2x-\frac{2}{5}\)
\(\Rightarrow4x-2x=\frac{18}{5}-\frac{2}{5}\Rightarrow2x=\frac{16}{5}\Rightarrow x=\frac{16}{5}:2=\frac{16}{10}=\frac{8}{5}\)
\(TH2:x< 1\Rightarrow|x-1|=-x+1\)
\(\Rightarrow2\left(-x+1+x-\frac{4}{5}\right)=2x-\frac{2}{5}\)
\(\Rightarrow2\left(1-\frac{4}{5}\right)=2x-\frac{2}{5}\Rightarrow2\cdot\frac{1}{5}=2x-\frac{2}{5}\)
\(\Rightarrow2x-\frac{2}{5}=\frac{2}{5}\Rightarrow2x=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}\Rightarrow x=\frac{4}{5}:2=\frac{4}{10}=\frac{2}{5}\)
Ta có : ( x+1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 30
( x + x + x +x +x ) + ( 1 + 2 + 3 + 4 + 5 ) = 30
x * 5 + 15 = 30
x * 5 = 30 - 15
x * 5 = 15
x = 15 : 5
x = 3
Vậy x = 3
Duyệt đi , chúc bạn học giỏi
(x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 4 + 5 + 6 + 7 + 8
x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 30
\(x\)x 5 + 15 = 30
\(x\)x 5 = 30 - 15 = 15
x = 15 : 5 = 3
Ta có :
\(4-\left|x-\frac{1}{5}\right|=-\frac{1}{2}\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=4-\left(-\frac{1}{2}\right)\)
\(\Leftrightarrow\)\(\left|x-\frac{1}{5}\right|=\frac{9}{2}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{1}{5}=\frac{9}{2}\\x-\frac{1}{5}=-\frac{9}{2}\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{9}{2}+\frac{1}{5}\\x=-\frac{9}{2}+\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{43}{10}\\x=-\frac{43}{10}\end{cases}}\)
Vậy \(x=\frac{43}{10}\) hoặc \(x=-\frac{43}{10}\)
Ta có: \(4-\left|x-\frac{1}{5}\right|=\frac{-1}{2}\)
\(\Rightarrow\left|x-\frac{1}{5}\right|=4-\frac{-1}{2}\)
\(\Rightarrow\left|x-\frac{1}{5}\right|=\frac{9}{2}\)
\(\Rightarrow x-\frac{1}{5}=\pm\frac{9}{2}\)
Nếu \(x-\frac{1}{5}=\frac{9}{2}\Rightarrow x=\frac{9}{2}+\frac{1}{5}=\frac{47}{10}\)
Nếu \(x-\frac{1}{5}=\frac{-9}{2}\Rightarrow x=\frac{-9}{2}+\frac{1}{5}=\frac{-43}{10}\)
Vậy \(x=\left\{\frac{47}{10};\frac{-43}{10}\right\}\)
Chúc bạn năm mới vui vẻ!