Tìm x biết:
I x - 1 + I x + 1 I = 2x - 3
Những câu hỏi liên quan
1) Tìm Min của các biểu thức :
a, A = | x + 1 | + | x + 2 | - 2x + 3
b, B = | 2x+3 | + | 1 - 2x |
c, C = | x - 1 | + 2| x - 2 |
a ) \(A=\left|x+1\right|+\left|x+2\right|-2x+3\ge2x+3-2x+3=6\)
Dấu " = " xảy ra khi \(\left(x+2\right)\left(x+1\right)\ge0\)
b )
\(B=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\)
Dấu " = " xảy ra khi \(\left(2x+3\right)\left(1-2x\right)\ge0\)
c )
\(C=\left|x-1\right|+\left|x-2\right|+\left|x-2\right|\ge\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1\)
Dấu " = " xảy ra khi \(x=2\)
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Tìm các số thực x, y thỏa mãn:
a) 2x + 1 + (1 – 2y)i = 2 – x + (3y – 2)i
b) 4x + 3 + (3y – 2)i = y +1 + (x – 3)i
c) x + 2y + (2x – y)i = 2x + y + (x + 2y)i
Bài 60: Tìm x; biết
a/ I x+1 I + I x + 2 I + ..... + I x + 100 I = 101x
b/ I x+ 1/1.2 I + I x + 1/2.3 I + ..... + I x + 1/99.100 I = 100x
c/I I 2x-1 I-1/2 I = 3/2
d/I I 3/2x - 2 I -5/2 I = 3/4
e/I x2 + 2018 I 2019x -1 I I = x2 + 2018
f/ I (x + 1/2 ) I 2x - 3/4 II = 2x -3/4
1, Tìm x thuộc Z biết :
a,2x - ( -5) = x- 4
b,- ( 2x+3) -x =2x -(- 3- 6)
c, (2x -3) -(2x +5) =6
d, I x I =3
e, I x -1 I-1 =3
g, Ix +1 =3 ( a thuộc Z )
các bạn làm mấy bài tìm x này theo bảng xét dấu hộ mk
a. I 5+2x I +2x=x-3
b.I x+1 I + I x+2 I=0
c.I 3x-1 I - I 2x+3 I = 0
d. I x-1 I -I 2-x I +x=0
Tìm x biết:
I x - 1 + I x + 1 I = 2x - 3
1. Tìm x
a). I x I + I x + 1 I + I x + 2 I + I x + 3 I + I x + 4 I = 5x
b). ( 2x - 5 ) - ( 3x - 7 ) = x+ 3
b) Theo bài ra , ta có :
(2x - 5) - (3x - 7) = x + 3
(=) 2x - 5 - 3x + 7 = x + 3
(=) -2x = 1
(=) x = -1/2
Vậy x = -1/2
Chúc bạn học tốt =))
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Tìm x biết
3(X-1/2) - 5(X-2/3) =3/2 x X
2X - 5/4 = (3 - 1/2) (X-1/2)
I x -1 I +1 = 2x - 3
Tìm X
a) (\(\dfrac{1}{4}\) - X) ( X + \(\dfrac{2}{5}\) ) = 0
b) I 2x + 1 I +\(\dfrac{2}{3}\) = 2
c) (2x - 3 )\(^2\) = 36
d) 7\(^x\) + 2 +2 x 7\(^x\) = 357
a: \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
b: \(\left|2x+1\right|+\dfrac{3}{2}=2\)
=>\(\left|2x+1\right|=\dfrac{1}{2}\)
=>\(\left[{}\begin{matrix}2x+1=\dfrac{1}{2}\\2x+1=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-\dfrac{1}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
c: (2x-3)2=36
=>\(\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
d: \(7^{x+2}+2\cdot7^x=357\)
=>\(7^x\cdot49+7^x\cdot2=357\)
=>\(7^x=7\)
=>x=1
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a) \(\left(\dfrac{1}{4}-x\right)\left(x+\dfrac{2}{5}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{4}-x=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
\(---\)
b) \(\left|2x+1\right| +\dfrac{2}{3}=2\)
\( \Rightarrow\left|2x+1\right|=2-\dfrac{2}{3}\)
\(\Rightarrow\left|2x+1\right|=\dfrac{4}{3}\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=\dfrac{4}{3}\\2x+1=-\dfrac{4}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{1}{3}\\2x=-\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{7}{6}\end{matrix}\right.\)
\(---\)
c) \(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=\left(\pm6\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=9\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
\(---\)
d) \(7^{x+2}+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot7^2+2\cdot7^x=357\)
\(\Rightarrow7^x\cdot\left(7^2+2\right)=357\)
\(\Rightarrow7^x\cdot\left(49+2\right)=357\)
\(\Rightarrow7^x\cdot51=357\)
\(\Rightarrow7^x=357:51\)
\(\Rightarrow7^x=7\)
\(\Rightarrow x=1\)
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