|x-3,2|+|2x-1/5|=x+3
a,|7 - 2x| + 7 = 2x
b,| 1 - x | = 4x + 1
c, | x - 1/3 | + 4/5 = | ( 3,2) + 2/5 |
d,| x - 7 | + 2x + 5 = 6
e, 3x - | 2x - 1 | = 2
a: \(\left|7-2x\right|+7=2x\)
=>\(\left|2x-7\right|+7=2x\)
=>\(\left|2x-7\right|=2x-7\)
=>2x-7>=0
=>\(x>=\dfrac{7}{2}\)
b: \(\left|1-x\right|=4x+1\)
=>\(\left|x-1\right|=4x+1\)
=>\(\left\{{}\begin{matrix}4x+1>=0\\\left(4x+1\right)^2=\left(x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1\right)^2-\left(x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left(4x+1-x+1\right)\left(4x+1+x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\5x\left(3x+2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{1}{4}\\\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
c: \(\left|x-\dfrac{1}{3}\right|+\dfrac{4}{5}=\left|3,2+\dfrac{2}{5}\right|\)
=>\(\left|x-\dfrac{1}{3}\right|=\dfrac{16}{5}+\dfrac{2}{5}-\dfrac{4}{5}=\dfrac{14}{5}\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{14}{5}\\x-\dfrac{1}{3}=-\dfrac{14}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{42+5}{15}=\dfrac{47}{15}\\x=-\dfrac{14}{5}+\dfrac{1}{3}=\dfrac{-42+5}{15}=-\dfrac{37}{15}\end{matrix}\right.\)
d: \(\left|x-7\right|+2x+5=6\)
=>\(\left|x-7\right|=6-2x-5=-2x+1\)
=>\(\left\{{}\begin{matrix}-2x+1>=0\\\left(-2x+1\right)^2=\left(x-7\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(2x-1+x-7\right)\left(2x-1-x+7\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left(3x-8\right)\left(x+6\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{8}{3}\left(loại\right)\\x=-6\left(nhận\right)\end{matrix}\right.\end{matrix}\right.\)
e: 3x-|2x-1|=2
=>|2x-1|=3x-2
=>\(\left\{{}\begin{matrix}3x-2>=0\\\left(3x-2\right)^2=\left(2x-1\right)^2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2\right)^2-\left(2x-1\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(3x-2-2x+1\right)\left(3x-2+2x-1\right)=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left(x-1\right)\left(5x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x-1=0\\5x-3=0\end{matrix}\right.\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=\dfrac{2}{3}\\\left[{}\begin{matrix}x=1\left(nhận\right)\\x=\dfrac{3}{5}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
Tĩm x, biết:
|x-3,2|+|2x-1/5|=x+3
Tìm x biết: \(\left|x-3,2\right|+\left|2x-\dfrac{1}{5}\right|=x+3\)
Bài này có 2 cách, cách 1 là xét 3 trường hợp, cách 2 là sử dụng phương pháp đánh giá. Trong bài này cách 2 ngắn hơn thì mình sẽ làm.
Điều kiện: x \(\ge\)0
Ta có: VT = |x - 3,2| + |2x - 0,2| = |3,2 - x| + |2x - 0,2| \(\ge\) |3,2 - x + 2x - 0,2| = |x + 3| = VP
Dấu "=" xảy ra <=> (3,2 - x)(2x - 0,2) \(\ge\) 0.
<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}3,2-x\ge0\\2x-0,2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le3,2\\x\ge0,1\end{matrix}\right.\Leftrightarrow0,1\le x\le3,2}}\\\left\{{}\begin{matrix}3,2-x\le0\\2x-0,2\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge3,2\\x\le0,1\end{matrix}\right.\Leftrightarrow x}\in\varphi}\end{matrix}\right.\)
tìm x biết a) |2x+3|=x+2
b) |x-1/3|+4/5=|(-3,2)+2/5
nhanh giùm đang cần gấp
ai nhanh mình tick cho
3,2 . x -(4/5 + 2/3) : 3 và 2/3 = 7/20
(4 và 1/2 - 2x) . 1 và 4/61 = 6 và 1/2
\(3,2x-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):3\dfrac{2}{3}=\dfrac{7}{20}\\ \Rightarrow\dfrac{16}{5}x-\dfrac{22}{15}:\dfrac{11}{3}=\dfrac{7}{20}\\ \Rightarrow\dfrac{16}{5}x-\dfrac{2}{5}=\dfrac{7}{20}\\ \Rightarrow\dfrac{16}{5}x=\dfrac{3}{4}\\ \Rightarrow x=\dfrac{15}{64}\)
\(\left(4\dfrac{1}{2}-2x\right).1\dfrac{4}{61}=6\dfrac{1}{2}\\ \Rightarrow\left(\dfrac{9}{2}-2x\right).\dfrac{65}{61}=\dfrac{13}{2}\\ \Rightarrow\dfrac{9}{2}-2x=\dfrac{61}{10}\\ \Rightarrow2x= -\dfrac{8}{5}\\ \Rightarrow x=-\dfrac{4}{5}\)
\(\text{|}x-3,2\text{|}+\text{|}2x-\frac{1}{5}\text{|}=x+3\)
Tìm x?
HELP MK NHA
Tìm x, sao cho:
|x+2|=|x-5|
|2x-3|=2-x
A=0,75-|x-3,2| có giá trị lớn nhất
B=2.|x+1,5|-3,2 có giá trị nhỏ nhất
Tìm x, biết: \(|x - 3,2| + \left|2x - \frac{1}{5}\right| = x + 3\)
Ai biết thì giúp mình với ạ! Mình đang cần gấp!
\(\left|x-3,2\right|+\left|2x-\frac{1}{5}\right|=x+3.\)
ĐK : \(x+3\ge0\Leftrightarrow x\ge-3\)
Th1 : \(x-3,2+2x-\frac{1}{5}=x+3\)
\(x-3,2+2x=x+\frac{16}{5}\)
\(x+2x=x+\frac{32}{5}\)
\(2x=\frac{32}{5}\)
\(\Leftrightarrow x=3,2\)(tm)
\(x-3,2+2x-\frac{1}{5}=3-x\)
\(x-3,2+2x=3-x+\frac{1}{5}\)
\(x-3,2+2x=\frac{16}{5}-x\)
\(x+2x=\frac{16}{5}-x+3,2\)
\(x+2x=\frac{32}{5}-x\)
\(2x=\frac{32}{5}-x-x\)
\(2x=\frac{32}{5}-2x\)
\(4x=\frac{32}{5}\)
\(x=1,6\)(tm)
Vậy \(x=1,6\)hoặc \(x=3,2\)
a) /3x-1/=/5-2x/
b) /2x-1/+x=2
c) A= 0,75-/x-3,2/ có giá trị lớn nhất
d) B=2./x+1,5/-3,2
Giúp mình với nha !!!
Mình cần gấp, ai làm đúng mình tick cho ạ!