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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
\(-4\left(x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=-3\) \(3\)
<=> \(-4\left(x^2-2x+1\right)+4x^2-1=-3\)
<=> \(-4x^2+8x-4+4x^2-1=-3\)
<=> \(8x-5=-3\)
<=> \(8x=2\)
<=> \(x=\frac{1}{4}\)
\(a,\left(x^2+2\right)\left(x^4-2x^2+4\right)=\left(x^2\right)^3+8=x^6+8\)
\(b,\left(x-\frac{1}{3}\right)\left(x^2+\frac{x}{3}+\frac{1}{9}\right)=x^3-\frac{1}{27}\)
\(c,\left(\frac{1}{2}-x\right)\left(\frac{1}{4}+\frac{1}{2}x+x^2\right)=\frac{1}{8}-x^3\)
\(d,\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)
\(e,\left(2x+1\right)\left(4x^2-2x+1\right)=8x^3+1\)
a) \(\left(x^2+2\right)\left(x^4-2x^2+4\right)=\left(x^2\right)^3+2^3=x^8+8\)
b) \(\left(x-\frac{1}{3}\right)\left(x^2+\frac{x}{3}+\frac{1}{9}\right)=[x^3-\left(\frac{1}{3}\right)^3]=x^3-\frac{1}{9}\)
c) \(\left(\frac{1}{2}-x\right)\left(\frac{1}{4}+\frac{1}{2}x+x^2\right)=[\left(\frac{1}{2}\right)^3-x^3]=\frac{1}{8}-x^3\)
d) \(\left(x^2+3\right)\left(x^4-3x^2+9\right)=\left(x^2\right)^3+3^3=x^8+27\)
e) \(\left(2x+1\right)\left(4x^2-2x+1\right)=\left(2x\right)^3+1^3=8x^3+1\)
<=> 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
a) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)
\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)
\(\Leftrightarrow-14x=14\)
\(\Leftrightarrow x=-1\)
b) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)
\(\Leftrightarrow17x=-34\Rightarrow x=-2\)
a,
\(\left(x-2\right)\left(2x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{2}\end{matrix}\right.\)
b,
\(\left(0,2x-3\right)\left(0,5x-8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}0,2x=3\\0,5x=8\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=15\\x=16\end{matrix}\right.\)
c,
\(2x\left(x-6\right)+3\left(x-6\right)=0\\ \Leftrightarrow\left(2x+3\right)\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1,5\\x=6\end{matrix}\right.\) (mình skip bớt cho đỡ lằng nhằng nhé :>)
d,
\(\left(x-1\right)\left(2x-4\right)\left(3x-9\right)=0\\ \Leftrightarrow6\left(x-1\right)\left(x-2\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=3\end{matrix}\right.\)
Chúc bạn học tốt nha
a, x-2=0\(\Leftrightarrow\) x=2
2x-5=0\(\Leftrightarrow\)2x=5\(\Leftrightarrow\)x=\(\dfrac{5}{2}\)
S=\(\left\{\dfrac{5}{2};2\right\}\)
b, 0.2x-3=0\(\Leftrightarrow\)0.2x=3\(\Leftrightarrow\)x=\(\dfrac{3}{0.2}\)
s=\(\left\{\dfrac{3}{0.2}\right\}\)
c, \(\Leftrightarrow\)(x-6)(2x+3)=0
\(\Leftrightarrow\)x-6=0\(\Leftrightarrow\)x=6
2x+3=0\(\Leftrightarrow\)2x=-3\(\Leftrightarrow\)x=\(\dfrac{-3}{2}\)
S=\(\left\{\dfrac{-3}{2};-3\right\}\)
D \(\Leftrightarrow\)x-1=0\(\Leftrightarrow\)x=1
2x-4=0\(\Leftrightarrow\)2x=4\(\Leftrightarrow\)x=2
3x-9=0\(\Leftrightarrow\)3x=9\(\Leftrightarrow\)x=3