|2x+1|+|x+3|=5x
Giúp mình với, Cảm ơn!!!
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a,=xy\left(5x-1\right)\\ b,=\left(x-2\right)\left(3x-5\right)\\ c,Sửa:x^2+2x+1-9y^2\\ =\left(x+1\right)^2-9y^2\\ =\left(x-3y+1\right)\left(x+3y+1\right)\\ d,=x\left(x-y\right)+5\left(x-y\right)=\left(x+5\right)\left(x-y\right)\)
Hình hiển thị bị lỗi rồi. Bạn nên gõ hẳn đề ra để được hỗ trợ tốt hơn nhé.
d) \(\left|2x-3\right|=x-3\)
TH1: \(\left|2x-3\right|=2x-3\) với \(2x-3\ge0\Leftrightarrow x\ge\dfrac{3}{2}\)
Pt trở thành:
\(2x-3=x-3\) (ĐK: \(x\ge\dfrac{3}{2}\) )
\(\Leftrightarrow2x-x=-3+3\)
\(\Leftrightarrow x=0\left(ktm\right)\)
TH2: \(\left|2x-3\right|=-\left(2x-3\right)\) với \(2x-3< 0\Leftrightarrow x< \dfrac{3}{2}\)
Pt trở thành:
\(-\left(2x-3\right)=x-3\)
\(\Leftrightarrow-2x+3=x-3\)
\(\Leftrightarrow-2x-x=-3-3\)
\(\Leftrightarrow-3x=-6\)
\(\Leftrightarrow x=-\dfrac{6}{-3}=2\left(ktm\right)\)
Vậy Pt vô nghiệm
\(\dfrac{-2}{3}\left(x-\dfrac{1}{4}\right)=\dfrac{1}{3}\left(2x-1\right)\)
\(\Leftrightarrow\dfrac{-2}{3}x+\dfrac{1}{6}=\dfrac{2}{3}x-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-2}{3}x-\dfrac{2}{3}x=\dfrac{-1}{3}-\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{-4}{3}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{3}{8}\)
Vậy \(x=\dfrac{3}{8}\)
\(\left|3x+5\right|=x+1\)
TH1: \(3x+5=x+1\left(x\ge-\dfrac{5}{3}\right)\)
\(\Rightarrow3x-x=1-5\)
\(\Rightarrow2x=-4\)
\(\Rightarrow x=-2\left(ktm\right)\)
TH2: \(3x-5=-\left(x+1\right)\left(x< -\dfrac{5}{3}\right)\)
\(\Rightarrow3x-5=-x-1\)
\(\Rightarrow3x+x=-1+5\)
\(\Rightarrow4x=4\)
\(\Rightarrow x=1\)
Vậy không có x thõa mãn
_______
\(\left|2x-3\right|=2x-3\)
\(\Rightarrow2x-3=2x-3\left(x\ge\dfrac{3}{2}\right)\)
\(\Rightarrow0=0\) (luôn đúng)
Nên mọi x đề thỏa mãn khi \(x\ge\dfrac{3}{2}\)
Vậy: ...
|3x + 5| = x + 1
TH1: x ≥log ) -5/3
(1) ⇒ 3x + 5 = x + 1
3x - x = 1 - 5
2x = -4
x = -2 (loại)
*) TH2: x < -5/3
(1) ⇒ 3x + 5 = -x - 1
3x + x = -1 - 5
4x = -6
x = -3/2 (loại)
Vậy không tìm được x thỏa mãn yêu cầu
--------
|2x - 3| = 2x - 3 (2)
*) TH1: x 3/2
(2) ⇒ 2x - 3 = 2x - 3
0x = 0 (luôn đúng với mọi x ≥ 3/2)
*) TH2: x < 3/2
(2) ⇒ 2x - 3 = 3 - 2x
2x + 2x = 3 + 3
4x = 6
x = 3/2 (loại)
Vậy x ≥ 3/2
\(\Leftrightarrow2x\left(x+5\right)-3\left(x-2\right)=7x+1\)
\(\Leftrightarrow2x^2+10x-3x+6-7x-1=0\)
\(\Leftrightarrow2x^2+5=0\)(vô lý)
ĐKXĐ:\(\left\{{}\begin{matrix}x\ne2\\x\ne-5\end{matrix}\right.\)
\(\dfrac{2x}{x-2}-\dfrac{3}{x+5}=\dfrac{7x+1}{x^2+3x-10}\\ \Leftrightarrow\dfrac{2x\left(x+5\right)}{\left(x+5\right)\left(x-2\right)}-\dfrac{3\left(x-2\right)}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x^2-2x+5x-10}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}=\dfrac{7x+1}{x\left(x-2\right)+5\left(x-2\right)}\\ \Leftrightarrow\dfrac{2x^2+10x}{\left(x+5\right)\left(x-2\right)}-\dfrac{3x-6}{\left(x+5\right)\left(x-2\right)}-\dfrac{7x+1}{\left(x+5\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2+10x-3x+6-7x-1}{\left(x+5\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{2x^2+5}{\left(x+5\right)\left(x-2\right)}=0\\ \Rightarrow2x^2+5=0\left(vô.lí\right)\)
Vậy pt vô nghiệm
ĐKXĐ: \(x\ge1\)
\(\Rightarrow\left(\sqrt{x-1}+\sqrt{2x+1}\right)^2=1\Leftrightarrow x-1+2x+1+2\sqrt{\left(x-1\right)\left(2x+1\right)}=1\Leftrightarrow3x+2\sqrt{2x^2-x-1}=1\) \(\Leftrightarrow2\sqrt{2x^2-x-1}=1-3x\Rightarrow\left(2\sqrt{2x^2-x-1}\right)^2=\left(1-3x\right)^2\Leftrightarrow8x^2-4x-4=9x^2-6x+1\) \(\Leftrightarrow x^2-2x+5=0\Leftrightarrow\left(x-1\right)^2+4=0\Leftrightarrow\left(x-1\right)^2=-4\) vô lí vì VT\(\ge0\) mà VP<0 \(\Rightarrow\) ko có x Vậy...
`@` `\text {Ans}`
`\downarrow`
\(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)
`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{75}{100}-\dfrac{3}{2}x=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{75}{100}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=-1\cdot3\\x=\dfrac{75}{100}\div\dfrac{3}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x={-3/2; 1/2}.`
a) \(4x^2-6x=2x\left(2x-3\right)\)
b) \(9x^4y^3+3x^2y^4=3x^2y^3\left(x^2+y\right)\)
c) \(x^3-2x^2+5x=x\left(x^2-2x+5\right)\)
2x + 1 + x + 3 = 5x
Suy ra 3x + 4 = 5x
4 = 5x - 3x
4 = 2x
x = 4 : 2 = 2
Đ/s : x = 2
giải
2x+1+x+3=5x
2x+4 =5x
4 =5x -3x
4 =2x
x =4:2
x = 2
học tốt nha