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14 tháng 2 2022
Bài 3:
a: =>4x+2=0
hay x=-1/2
b: \(\Leftrightarrow\left[{}\begin{matrix}5x-10=0\\6x+2=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;-\dfrac{1}{3}\right\}\)
Bài 4:
d: \(\text{Δ}=\left(-6\right)^2-4\cdot2\cdot1=36-8=28>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{7}}{4}=\dfrac{3-\sqrt{7}}{2}\\x_2=\dfrac{3+\sqrt{7}}{2}\end{matrix}\right.\)
14 tháng 2 2022
Bạn đăng tách ra nhé
Bài 1 : a, \(15-8x=9-5x\Leftrightarrow-3x=-6\Leftrightarrow x=2\)
b, \(3+2x=5x+2\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)
c, \(5-x+6=12-8x\Leftrightarrow1=7x\Leftrightarrow x=\dfrac{1}{7}\)
d, \(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\Leftrightarrow8x=-16\Leftrightarrow x=-2\)
T
5 tháng 9 2023
4. \(x^2-3x+xy-3y=0\)
\(\Leftrightarrow x\left(x-3\right)+y\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-y\end{matrix}\right.\)
5. \(x^2-8x-3x+24=0\)
\(\Leftrightarrow x\left(x-8\right)-3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\)
6. \(\left(x-2\right)^2-5\left(2-x\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
7. \(3x\left(x-1\right)-x^2+2x-1=0\)
\(\Leftrightarrow3x\left(x-1\right)-\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[3x-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
8. \(x^2\left(x-3\right)+18-6x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\pm\sqrt{6}\end{matrix}\right.\)
10. \(\left(x-5\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[\left(x-5\right)-\left(x-2\right)\right]\left[\left(x-5\right)+\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-5-x+2\right)\left(x-5+x-2\right)=0\)
\(\Leftrightarrow-3\left(2x-7\right)=0\)
\(\Leftrightarrow2x-7=0\)
\(\Leftrightarrow x=\dfrac{7}{2}\)
12. \(x^2\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\)
14. \(3x^2-7x-10=0\)
\(\Leftrightarrow3x^2+3x-10x-10=0\)
\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)
#Urushi
5 tháng 9 2023
4: x^2-3x+xy-3y=0
=>x(x-3)+y(x-3)=0
=>(x-3)(x+y)=0
=>x=3 và x+y=0
=>x=3 và y=-3
6: (x-2)^2-5(2-x)=0
=>(x-2)^2+5(x-2)=0
=>(x-2)(x-2+5)=0
=>(x-2)(x+3)=0
=>x=-3 hoặc x=2
8: x^2(x-3)+18-6x=0
=>x^2(x-3)-6(x-3)=0
=>(x-3)(x^2-6)=0
=>x=3 hoặc \(x=\pm\sqrt{6}\)
10: (x-5)^2-(x-2)^2=0
=>(x-5-x+2)(x-5+x-2)=0
=>-3(2x-7)=0
=>2x-7=0
=>x=7/2
12: x^2(x-3)-4x+12=0
=>x^2(x-3)-4(x-3)=0
=>(x-3)(x^2-4)=0
=>(x-3)(x-2)(x+2)=0
=>\(x\in\left\{3;2;-2\right\}\)
14: 3x^2-7x-10=0
=>3x^2-10x+3x-10=0
=>(3x-10)(x+1)=0
=>x=10/3 hoặc x=-1
??????/