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\(\left(2x-2\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-2\right)-\left(2x-1\right)\right]\cdot\left[\left(2x-2\right)+\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(2x-2-2x+1\right)\cdot\left(2x-2+2x-1\right)=0\)
\(\Leftrightarrow\left(2x-2x-2+1\right)\cdot\left(2x+2x-2-1\right)=0\)
\(\Leftrightarrow\left(-1\right)\cdot\left(4x-3\right)=0\)
\(\Leftrightarrow4x-3=0\div\left(-1\right)\)
\(\Leftrightarrow4x-3=0\)
\(\Leftrightarrow4x=3\)
\(\Leftrightarrow x=\frac{3}{4}\)
Vậy \(x=\frac{3}{4}\)
\(\left(2x-2\right)^2-\left(2x-1\right)^2=0\)
\(\left[2x-2-\left(2x-1\right)\right]\left[2x-2+\left(2x-1\right)\right]=0\)
\(\left(2x-2-2x+1\right)\left(2x-2+2x-1\right)=0\)
\(-1\left(4x-3\right)=0\)
\(-4x+3=0\)
\(-4x=-3\)
\(x=\frac{3}{4}\)
Ta có : (x + 1)2 - (x + 2)(x - 2) = 0
<=> (x + 1)2 - (x2 - 22) = 0
<=> x2 + 2x + 1 - x2 + 4 = 0
<=> 2x + 5 = 0
=> 2x = -5
=> x = \(-\frac{5}{2}\)
\(\Leftrightarrow\frac{1}{x^2+5x+6}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+x}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x^2+5x+6}+\frac{1}{x^2+3x+2}+\frac{1}{x^2+x}-\frac{3}{10}=0\)
\(\Leftrightarrow-\frac{3\left(x^2+3x-10\right)}{10x\left(x+3\right)}=0\)
\(\Leftrightarrow3\left(x^2+3x-10\right)=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x^2-2x+5x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow x-2=0\)hoặc\(x+5=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+3}=\frac{3}{10}\Leftrightarrow\frac{\left(x+3\right)-x}{x\left(x+3\right)}=\frac{3}{10}\Leftrightarrow\frac{3}{x\left(x+3\right)}=\frac{3}{10}\)
<=>x(x+3)=10 <=> x2+3x=10 <=> x2+3x-10=0
<=>-(x2-3x+10)=0
<=>x2-3x+10=0
<=>x2-2.x.\(\frac{3}{2}\)+ \(\left(\frac{3}{2}\right)^2+\frac{31}{4}\)=0
<=> \(\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\)=0
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{31}{4}\ge\frac{31}{4}>0\) (với mọi x)
=>PT vô nghiệm
Em kiểm tra lại đề bài nhé \(\frac{2}{x-y}\)hay \(\frac{2}{x-2}\)
a/ \(\orbr{\begin{cases}x-2=0\\2x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{5}{2}\end{cases}}\)
\(a,\left(x-2\right)\left(2x-5\right)=0.\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\2x=5\Leftrightarrow x=\frac{5}{2}\end{cases}}}\)
Vậy ....
\(b,\left(0,2x-3\right)\left(0,5x-8\right)=0\left(\text{Mạo muội sửa đề nha 0,5 thành 0,5x}\right)\)
\(\Leftrightarrow\orbr{\begin{cases}0,2x-3=0\\0,5x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}0,2x=3\\0,5x=8\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=15\\x=16\end{cases}}\)
Vậy ... ( có j sai thì bỏ qua cho)
\(c,2x\left(x-6\right)+3\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\2x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=6\\2x=-3\Leftrightarrow x=-\frac{3}{2}\end{cases}}}\)
Vậy ...
\(d,\left(x-1\right)\left(2x-4\right)\left(3x-9\right)=0\)
\(\Leftrightarrow2.3\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\)
( ko có ngoặc vuông 3 cái nên mk trình bày kiểu này)
+ TH1:
x-1=0 <=> x= 1
+ TH2:
x-2=0 <=> x=2
+TH3:
x-3 = 0 <=> x = 3
\(\left(2x^3+x^2+10x+30\right):\left(2x+1\right)\)
\(=2x^3:\left(2x+1\right)+x^2:\left(2x+1\right)+10x:\left(2x+1\right)+30:\left(2x+1\right)\)
\(=2x^3:2x+2x^3:1+x^2:2x+x^2:1+10x:2x+10x:1+30:2x+30:1\)
\(=x^2+2x^3+\dfrac{1}{2}x+x^2+5+10x+15x+30\)
\(=2x^3+2x^2+\dfrac{51}{2}x+35\)
<=> 4(x^2 + 2x + 1) + 4x^2 - 4x +1 - 8(x^2 - 1) = 11
<=> 4x^2 + 8x + 4 + 4x^2 - 4x +1 - 8x^2 +8 - 11 = 0
<=> 4x + 2 = 0
<=> x = - 1/2
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
\(4\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)=11\)
\(4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(4x+13=11\)
\(4x=-2\)
\(x=-\frac{1}{2}\)