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TN
13 tháng 3 2020
\(a.\frac{x-6}{x-4}=\frac{x}{x-2}\\\Leftrightarrow \frac{\left(x-6\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=\frac{x\left(x-4\right)}{\left(x-4\right)\left(x-2\right)}\\\Leftrightarrow \left(x-6\right)\left(x-2\right)=x\left(x-4\right)\\\Leftrightarrow \left(x-6\right)\left(x-2\right)-x\left(x-4\right)=0\\ \Leftrightarrow x^2-2x-6x+12-x^2+4x=0\\\Leftrightarrow -4x+12=0\\\Leftrightarrow -4x=-12\\ \Leftrightarrow x=3\)
\(b.1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\\ \Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)=0\\ \Leftrightarrow x^2-x-2x+3+2x^2-2x-5x+5-3x^2+6x+5x-10=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \)
ST
2
21 tháng 4 2020
\(\frac{x+5}{x^2-5x}-\frac{x+25}{2x^2-50}=\frac{x-5}{2x^2+10x}\) (α)
ĐKXĐ : \(x\ne0;x\ne\pm5\)
Với đk trên, ta có :
(α) ⇔ \(\frac{2\left(x+5\right)^2}{2x\left(x^2-25\right)}-\frac{x^2+25x}{2x\left(x^2-25\right)}=\frac{\left(x-5\right)^2}{2x\left(x^2-25\right)}\)
⇔ \(2\left(x^2+10x+25\right)-x^2-25x=x^2-10x+25\)
⇔ \(2x^2+20x+50-x^2-25x=x^2-10x+25\)
⇔ \(2x^2-x^2-x^2+20x-25x+10x=25-50\)
⇔ \(5x=-25\)
⇔ \(x=-5\) (loại)
Vậy : \(S=\varnothing\)
28 tháng 10 2018
\(I=3\left(x^2-\dfrac{5}{3}x+1\right)\)
\(I=3\left(x^2-2.x.\dfrac{5}{6}+\left(\dfrac{5}{6}\right)^2-\left(\dfrac{5}{6}\right)^2+1\right)\)
\(I=3\left[\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{36}\right]\)
\(I=3\left(x-\dfrac{5}{6}\right)^2+\dfrac{11}{12}\)
T
28 tháng 10 2018
mình ra là \(\dfrac{11}{36}\)mà bn
bn coi lại đi
I=3x2-5x+3
I=3(x2-\(\dfrac{5}{3}\)x+1)
I=3[x2-2.x.\(\dfrac{5}{3}\)+\(\left(\dfrac{5}{6}\right)^2\)-\(\left(\dfrac{5}{6}\right)^2\)+1]
I=3(x-\(\dfrac{5}{3}\))2+\(\dfrac{11}{36}\)
I=3(x-\(\dfrac{5}{3}\))2+\(\dfrac{11}{36}\)≥\(\dfrac{11}{36}\)
vậy Min I= \(\dfrac{11}{36}\)khi x =\(\dfrac{5}{3}\)
Theo mik nghĩ là vậy á
CHÚC BN HỌC TỐT
12 tháng 8 2020
Bài 1:
a) Ta có: \(\left(x^2-2x+1\right):\left(x-1\right)\)
\(=\left(x-1\right)^2:\left(x-1\right)\)
=x-1
b) Ta có: \(\left(x^3+1\right):\left(x^2-x+1\right)\)
\(=\frac{\left(x+1\right)\left(x^2-x+1\right)}{x^2-x+1}=x+1\)
c) Ta có: \(\left(x^3-x^2-5x-3\right):\left(x-3\right)\)
\(=\frac{x^3-3x^2+2x^2-6x+x-3}{x-3}\)
\(=\frac{x^2\left(x-3\right)+2x\left(x-3\right)+\left(x-3\right)}{\left(x-3\right)}\)
\(=\frac{\left(x-3\right)\left(x^2+2x+1\right)}{\left(x-3\right)}\)
\(=\left(x+1\right)^2\)
d) Ta có: \(\left(x^4+x^3-6x^2-5x+5\right):\left(x^2+x-1\right)\)
\(=\frac{x^4+x^3-x^2-5x^2-5x+5}{x^2+x-1}\)
\(=\frac{x^2\left(x^2+x-1\right)-5\left(x^2+x-1\right)}{x^2+x-1}\)
\(=\frac{\left(x^2+x-1\right)\left(x^2-5\right)}{x^2+x-1}\)
\(=x^2-5\)
BL
10 tháng 3 2020
1. \(1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\)
\(\Rightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\)
\(\Rightarrow x^2-x-2x+2+2x^2-2x-5x+5-3x^2+6x+5x-10=0\)
\(\Rightarrow x-3=0\Rightarrow x=3\)
2. \(\frac{x-3}{x-2}-\frac{x-2}{x-4}=3\frac{1}{5}\)
\(\Rightarrow\frac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}-\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}-\frac{16}{5}=0\)
\(\Rightarrow x^2-4x-3x+12-x^2+4x-4-16=0\)
\(\Rightarrow-3x-8=0\Rightarrow x=\frac{-8}{3}\)
3. \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{x^2-4}\)
\(\Rightarrow\frac{\left(x-2\right)^2}{x^2-4}-\frac{3\left(x+2\right)}{x^2-4}-\frac{2\left(x-11\right)}{x^2-4}=0\)
\(\Rightarrow x^2-4x+4-3x-6-2x+22=0\)
\(\Rightarrow x^2-9x+20=0\)
\(\Rightarrow x^2-4x-5x+20=0\)
\(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
BL
27 tháng 2 2020
\(a.\frac{7x-3}{x-1}=\frac{3}{2}\)
\(\Leftrightarrow\frac{7x-3}{x-1}-\frac{3}{2}=0\)
\(\Leftrightarrow\frac{2\left(7x-3\right)}{2.\left(x-1\right)}-\frac{3\left(x-1\right)}{2\left(x-1\right)}=0\)
\(\Leftrightarrow\frac{14x-6-3x+3}{2\left(x-1\right)}=0\)
\(\Leftrightarrow11x-3=0\)
\(\Leftrightarrow x=\frac{3}{11}\)
\(b.\frac{2\left(3-7x\right)}{1+x}=\frac{1}{2}\)
\(\Leftrightarrow\frac{6-14x}{1+x}-\frac{1}{2}=0\)
\(\Leftrightarrow\frac{2\left(6-14x\right)}{2\left(1+x\right)}-\frac{1+x}{2\left(1+x\right)}=0\)
\(\Leftrightarrow\frac{12-28x-1-x}{2\left(1+x\right)}=0\)
\(\Leftrightarrow11-29x=0\)
\(\Leftrightarrow x=\frac{11}{29}\)
\(c.\frac{1}{x-2}+3=\frac{3-x}{x-2}\)
\(\Leftrightarrow\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}-\frac{3-x}{x-2}=0\)
\(\Leftrightarrow\frac{1+3x-6-3+x}{x-2}=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow x=2\)
\(d.\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
\(\Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}-\frac{20}{x^2-25}=0\)
\(\Leftrightarrow\frac{x^2+10x+25-x^2+10x-25-20}{x^2-25}=0\)
\(\Leftrightarrow20x-20=0\)
\(\Leftrightarrow x=10\)
31 tháng 12 2022
a: \(\Leftrightarrow4\left(6-x\right)-3x=6\left(2x+3\right)-12\)
=>24-4x-3x=12x+18-12
=>12x+6=-7x+24
=>19x=18
=>x=18/19
b: \(\Leftrightarrow-210x-6\left(x-3\right)-15x=30x+10\left(2x+1\right)\)
=>-225x-6x+18=30x+20x+10
=>-231x+18-50x-10=0
=>-281x=-8
=>x=8/281
c: \(\Leftrightarrow36-2\left(x+3\right)=-4x+1-x\)
=>36-2x-6=-5x+1
=>3x=1+6-36=5-36=-31
=>x=-31/3
d: \(\Leftrightarrow-30\left(x-3\right)+10\left(2x-7\right)=6\left(6-x\right)\)
=>-30x+90+20x-70=36-6x
=>-10x+20=36-6x
=>-4x=16
=>x=-4
\(x^2=5\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)