2/3+2x=11/2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Thay \(x = 2\) vào phương trình \(\sqrt { - 2{x^2} - 2x + 11} = \sqrt { - {x^2} + 3} \) ta thấy không thỏa mãn vì dưới dấu căn là \( - 1\) không thỏa mãn
Vậy \(x = 2\) không là nghiệm của phương trình do đó lời giải như trên là sai.
a: \(3\left(x-3\right)-6x=0\)
=>\(3x-9-6x=0\)
=>-3x-9=0
=>3x+9=0
=>3x=-9
=>\(x=-\dfrac{9}{3}=-3\)
b: Đề thiếu vế phải rồi bạn
c: \(2\left(x-3\right)+3x=9\)
=>2x-6+3x=9
=>5x-6=9
=>5x=6+9=15
=>x=15/5=3
d: \(x\left(x-11\right)+2\left(x-11\right)=0\)
=>\(\left(x-11\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)
e: \(x\left(x+2\right)+8=x^2\)
=>\(x^2+2x+8=x^2\)
=>2x+8=0
=>2x=-8
=>x=-8/2=-4
f: \(8\left(x+1\right)+2x=-2\)
=>\(8x+8+2x=-2\)
=>10x=-2-8=-10
=>\(x=-\dfrac{10}{10}=-1\)
g: 12-3(x+2)=0
=>3(x+2)=12
=>x+2=12/3=4
=>x=4-2=2
b: \(B=2x\left(x-3\right)-\left(2x-2\right)\left(x-2\right)\)
\(=2x^2-6x-2x^2+4x+2x-4\)
=-4
\(\dfrac{2x-3}{2}>\dfrac{8x-11}{6}\)
\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{6}>\dfrac{8x-11}{6}\)
\(\Leftrightarrow3\left(2x-3\right)>8x-11\)
\(\Leftrightarrow6x-9>8x-11\)
\(\Leftrightarrow-2x>-2\)
\(\Leftrightarrow x< 1\)
Vậy \(S=\left\{x|x< 1\right\}\)
\(2x-3\le8x-11\)
\(\Leftrightarrow-6x\le-8\)
\(\Leftrightarrow x\ge\dfrac{8}{6}\)
Vậy \(S=\left\{x|x\ge\dfrac{8}{6}\right\}\)
1.|x+1|+|3−2x|=|3x−2|
mà |3−2x|=|3x−2| nên |x+1|=0 => x+1=0 =>x=-1
\(4\left(x+1\right)\left(-x+2\right)+\left(2x-1\right)\left(2x+3\right)=-11\)
\(\text{⇔}-4x^2+4x+8+4x^2+4x-3=-11\)
\(\text{⇔}8x+5=-11\)
\(\text{⇔}8x=-16\)
\(\text{⇔}x=-2\)
Vậy: \(x=-2\)
==========
\(\left(2x+4\right)\left(3x+1\right)\left(x-2\right)-\left(-3x^2+1\right)\left(-2x+\dfrac{2}{3}\right)=-\dfrac{26}{3}\)
\(\text{⇔}6x^3+2x^2-24x-8-6x^3-2x^2-2x+\dfrac{2}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x-\dfrac{22}{3}=-\dfrac{26}{3}\)
\(\text{⇔}-26x=-\dfrac{4}{3}\)
\(\text{⇔}x=\dfrac{2}{39}\)
e) \(\left(x+3\right)^3=\left(2x\right)^3\)
\(\Rightarrow x+3=2x\)
\(\Rightarrow2x-x=3\)
\(\Rightarrow x=3\)
f) \(\left(5-x\right)^5=32\)
\(\Rightarrow\left(5-x\right)^5=2^5\)
\(\Rightarrow5-x=2\)
\(\Rightarrow x=5-2\)
\(\Rightarrow x=3\)
g) \(\left(5x-6\right)^3=64\)
\(\Rightarrow\left(5x-6\right)^3=4^3\)
\(\Rightarrow5x-6=4\)
\(\Rightarrow5x=4+6\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=\dfrac{10}{2}\)
\(\Rightarrow x=5\)
h) \(5\cdot9^x=405\)
\(\Rightarrow9^x=\dfrac{405}{5}\)
\(\Rightarrow9^x=81\)
\(\Rightarrow9^x=9^2\)
\(\Rightarrow x=2\)
i) \(11^5:11^{n-2}=11^5\)
\(\Rightarrow11^{n-2}=11^5:11^5\)
\(\Rightarrow11^{n-2}=1\)
\(\Rightarrow11^{n-2}=11^0\)
\(\Rightarrow n-2=0\)
\(\Rightarrow n=2\)
k) \(\left(3x\right)^3=\left(2x+1\right)^3\)
\(\Rightarrow3x=2x+1\)
\(\Rightarrow3x-2x=1\)
\(\Rightarrow x=1\)
\(\frac{2}{3}+2x=\frac{11}{2}\)
\(\Rightarrow2x=\frac{11}{2}-\frac{2}{3}\)
\(\Rightarrow2x=\frac{33}{6}-\frac{4}{6}\)
\(\Rightarrow2x=\frac{29}{6}\)
\(\Rightarrow x=\frac{29}{12}\)