Tìm x:
1/2×(3x +1)-2:3=4
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d: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)
b) \(\Rightarrow x^2-13x+22=0\)
\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)
c) \(\Rightarrow x^2-3x-10=0\)
\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
a) 3/2.|x - 5/3| - 4/5 = 4/3.|x - 5/3| + 1
<=> 3/2.|x - 5/3| = 4/3.|x - 5/3| + 1 + 4/5
<=> 3/2.|x - 5/3| = 9/5 + 4|x - 5/3|/3
<=> 3/2.|x - 5/3| - 4.|x - 5/3|/3 = 9/5
<=> |x - 5/3|/6 = 9/5
<=> |x - 5/3| = 9/5.6
<=> |x - 5/3| = 54/5
<=> x - 5/3 = 54/5 hoặc x - 5/3 = -54/5
x = 54/5 + 5/3 x = -54/5 - 5/3
x = 187/15 x = -137/15
b) 2.|3x + 1| = 1/3.|3x + 1| + 5
<=> 2.|3x + 1| - 1/3.|3x + 1| = 5
<=> 5/3.|3x + 1| = 5
<=> 5.|3x + 1| = 5.3
<=> 5.|3x + 1| = 15
<=> |3x + 1| = 15 : 5
<=> |3x + 1| = 3
3x + 1 = 3 hoặc 3x + 1 = -3
3x = 3 - 1 3x = -3 - 1
3x = 2 3x = -4
x = 2/3 x = -4/3
=> x = 2/3 hoặc x = -4/3
c) làm tương tự câu a) mình hơi lời
Làm câu c) cho
\(\frac{1}{4}-\frac{5}{2}\left|3x-\frac{1}{5}\right|=\frac{2}{3}\left|3x-\frac{1}{5}\right|-\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{4}+\frac{2}{3}=\frac{2}{3}\left|3x-\frac{1}{5}\right|+\frac{5}{2}\left|3x-\frac{1}{5}\right|\)
\(\Leftrightarrow\frac{3}{12}+\frac{8}{12}=\left|3x-\frac{1}{5}\right|\left(\frac{2}{3}+\frac{5}{2}\right)\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|\left(\frac{4}{6}+\frac{15}{6}\right)=\frac{11}{12}\)
\(\Leftrightarrow\frac{19}{6}\left|3x-\frac{1}{5}\right|=\frac{11}{12}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{12}.\frac{6}{19}\)
\(\Leftrightarrow\left|3x-\frac{1}{5}\right|=\frac{11}{38}\)
\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{1}{5}=\frac{11}{38}\\3x-\frac{1}{5}=-\frac{11}{38}\end{cases}}\)
Giải tiếp nha
Tìm min:
$F=3x^2+x-2=3(x^2+\frac{x}{3})-2$
$=3[x^2+\frac{x}{3}+(\frac{1}{6})^2]-\frac{25}{12}$
$=3(x+\frac{1}{6})^2-\frac{25}{12}\geq \frac{-25}{12}$
Vậy $F_{\min}=\frac{-25}{12}$. Giá trị này đạt tại $x+\frac{1}{6}=0$
$\Leftrightarrow x=\frac{-1}{6}$
Tìm min
$G=4x^2+2x-1=(2x)^2+2.2x.\frac{1}{2}+(\frac{1}{2})^2-\frac{5}{4}$
$=(2x+\frac{1}{2})^2-\frac{5}{4}\geq 0-\frac{5}{4}=\frac{-5}{4}$ (do $(2x+\frac{1}{2})^2\geq 0$ với mọi $x$)
Vậy $G_{\min}=\frac{-5}{4}$. Giá trị này đạt tại $2x+\frac{1}{2}=0$
$\Leftrightarrow x=\frac{-1}{4}$
nhìn cái đề con hơi bị ''sốc'' , thế này ạ ???
Sửa đề \(4+\frac{1}{3}x\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}x\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)
\(4+\frac{1}{3}x\left(-\frac{1}{3}\right)\le x\le\frac{2}{3}x\left(-\frac{11}{12}\right)\)
\(4-\frac{1}{9}x\le x\le-\frac{11}{18}x\)
1.a) có: \(|x-\frac{3}{2}|,|x+1|,\left|x-2\right|\ge0\Rightarrow4x\ge0\Rightarrow x\ge0\)
\(x\ge0\Rightarrow x-\frac{3}{2}\ge\frac{-3}{2}\Rightarrow\left|x-\frac{3}{2}\right|\ge\left|\frac{-3}{2}\right|=\frac{3}{2}\Rightarrow\left|x-\frac{3}{2}\right|=x-\frac{3}{2}\)
cmtt: \(|x-2|=x-2\)
\(\Rightarrow3x-\frac{3}{2}+1-2=4x\)
\(\Rightarrow3x-\frac{5}{2}=4x\)
\(\Rightarrow x=\frac{-5}{2}\left(ko,t/m\right)\)
2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
\(\frac{1}{2}\times\left(3x+1\right)-2:3=4\)
\(\frac{1}{2}\times\left(3x+1\right)-\frac{2}{3}=4\)
\(\frac{1}{2}\times\left(3x+1\right)-\frac{1}{2}\times\frac{4}{3}=4\)
\(\frac{1}{2}\times\left(3x+1-\frac{4}{3}\right)=4\)
\(3x+\frac{-1}{3}=4:\frac{1}{2}\)
\(3x+\frac{-1}{3}=8\)
\(3x=\frac{24}{3}-\left(\frac{-1}{3}\right)\)
\(3x=\frac{25}{3}\)
\(x=\frac{25}{3}:3\)
\(x=\frac{25}{9}\)