Tìm x :
x2-7x+10=0
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1
AT
11 tháng 7 2021
a) \(\left(x^2-3x\right)\left(x^2+7x+10\right)=216\Rightarrow x\left(x-3\right)\left(x+2\right)\left(x+5\right)=216\)
\(\Rightarrow x\left(x+2\right)\left(x-3\right)\left(x+5\right)=216\Rightarrow\left(x^2+2x\right)\left(x^2+2x-15\right)=216\)
Đặt \(t=x^2+2x\Rightarrow\) pt trở thành \(t\left(t-15\right)=216\Rightarrow t^2-15t-216=0\)
\(\Rightarrow\left(t+9\right)\left(t-24\right)=0\Rightarrow\left[{}\begin{matrix}t=-9\\t=24\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+2x=-9\\x^2+2x=24\end{matrix}\right.\)
\(TH_1:x^2+2x=-9\Rightarrow x^2+2x+9=0\Rightarrow\left(x+1\right)^2+8=0\) (vô lý)
\(TH_2:x^2+2x=24\Rightarrow x^2+2x-24=0\Rightarrow\left(x-4\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
b) \(\left(2x^2-7x+3\right)\left(2x^2+x-3\right)+9=0\)
\(\Rightarrow\left(x-3\right)\left(2x-1\right)\left(x-1\right)\left(2x+3\right)+9=0\)
\(\Rightarrow\left(x-3\right)\left(2x+3\right)\left(x-1\right)\left(2x-1\right)+9=0\)
\(\Rightarrow\left(2x^2-3x-9\right)\left(2x^2-3x+1\right)+9=0\)
Đặt \(t=2x^2-3x-9\Rightarrow\) pt trở thành \(t\left(t+10\right)+9=0\)
\(\Rightarrow t^2+10t+9=0\Rightarrow\left(t+1\right)\left(t+9\right)=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-9\end{matrix}\right.\)
\(TH_1:t=-1\Rightarrow2x^2-3x-9=-1\Rightarrow2x^2-3x-8=0\)
\(\Delta=\left(-3\right)^2-4\left(-8\right).2=73\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-\sqrt{73}}{4}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{73}}{4}\end{matrix}\right.\)
\(TH_2:t=-9\Rightarrow2x^2-3x-9=-9\Rightarrow2x^2-3x=0\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
18 tháng 8 2021
a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)
18 tháng 8 2021
a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)
\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{5}\end{matrix}\right.\)
b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)^2\cdot\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)
c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+5-2\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-1\end{matrix}\right.\)
26 tháng 10 2021
a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
2 tháng 11 2021
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
BE
1
AH
Akai Haruma
Giáo viên
6 tháng 2
Lời giải:
$x^2-7x-8=0$
$\Leftrightarrow (x^2+x)-(8x+8)=0$
$\Leftrightarrow x(x+1)-8(x+1)=0$
$\Leftrightarrow (x+1)(x-8)=0$
$\Rightarrow x+1=0$ hoặc $x-8=0$
$\Rightarrow x=-1$ hoặc $x=8$
22 tháng 3 2020
Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0
=x(x+3)+2(x+3)=(x+2)(x+3)=0
Dễ rồi
2)\(x^2-x-6=0=x^2-3x+2x-6=0\)
=x(x-3)+2(x-3)=0
=(x+2)(x-3)=0
Dễ rồi
3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)
Vì \(x^2+1>0\)
=>\(\left(x+2\right)^2=0\)
Dễ rồi
4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0
=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)
=>x+1=0
=>..................
5)\(x^2-7x+6=x^2-6x-x+6\) =0
=x(x-6)-(x-6)=0
=(x-1)(x-6)=0
=>.....
6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0
=2x(x+1)-5(x+1)=0
=(2x-5)(x+1)=0
7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0
Dễ rồi
Nghỉ đã hôm sau làm mệt
6 tháng 2 2022
\(PT\Leftrightarrow2\left(x+7\right)-x\left(x+7\right)=0\)
\(\Leftrightarrow\left(x+7\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=2\end{matrix}\right.\)
Vậy: \(S=\left\{-7;2\right\}\)
3 tháng 3 2022
a: (2x-10)(5x+25)=0
=>2x-10=0 hoặc 5x+25=0
=>x=5 hoặc x=-5
b: (x+15)(x-2)=0
=>x+15=0 hoặc x-2=0
=>x=-15 hoặc x=2
c: =>x(x-7)=0
=>x=0 hoặc x=7
\(x^2-7x+10=0\)
\(\left(x^2-2x\right)-\left(5x-10\right)=0\)
\(x.\left(x-2\right)-5\left(x-2\right)=0\)
\(\left(x-2\right)\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
Tham khảo nhé~
\(x^2-7x+10=0\)
\(\Leftrightarrow x^2-5x-2x+10=0\)
\(\Leftrightarrow x\left(x-5\right)-2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}}\)