giải pt:
a) (x2-3x)(x2+7x+10)=216
b) (2x2-7x+3)(2x2+x-3)+9=0
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Hà Quang Minh
Giáo viên
19 tháng 9 2023
a) (x2 – x) . (2x2 – x – 10)
= x2 . (2x2 – x – 10) – x. (2x2 – x – 10)
= x2 . 2x2 + x2 . (-x) + x2 .(-10) – [ x. 2x2 + x. (-x) + x. (-10)]
= 2x4 – x3 - 10x2 – (2x3 – x2 – 10x)
= 2x4 – x3 - 10x2 – 2x3 + x2 + 10x
= 2x4 + (– x3 – 2x3 ) + (-10x2 + x2 )+ 10x
= 2x4 – 3x3 - 9x2 + 10x
b) (0,2x2 – 3x) . 5(x2 -7x + 3)
= (0,2x2 . 5 – 3x . 5) . (x2 -7x + 3)
= (x2 – 15x). (x2 -7x + 3)
= x2 . (x2 -7x + 3) – 15x. (x2 -7x + 3)
= x2 . x2 + x2 . (-7x) + x2 . 3 – [ 15x3 + 15x.(-7x) + 15x.3]
= x4 – 7x3 + 3x2 – (15x3 – 105x2 + 45x)
= x4 – 7x3 + 3x2 – 15x3 + 105x2 – 45x
= x4 +(– 7x3 – 15x3 )+ (3x2 + 105x2) – 45x
= x4 – 22x3 + 108x2 – 45x
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
23 tháng 10 2021
\(a,\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow x^3-27-x^3+4x=1\\ \Leftrightarrow4x=28\Leftrightarrow x=7\\ c,\Leftrightarrow4x^2-4x-8=0\\ \Leftrightarrow x^2-x-2=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ d,\Leftrightarrow2x^2+6x+x+3=0\\ \Leftrightarrow\left(x+3\right)\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
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 \(t\left(t+10\right)+9=0\)
\(\Rightarrow t^2+10t+9=0\Rightarrow\left(t+1\right)\left(t+9\right)=0\Rightarrow\left[{}\begin{matrix}t=-1\\t=-9\end{matrix}\right.\)
\(TH_1:t=-1\Rightarrow2x^2-3x-9=-1\Rightarrow2x^2-3x-8=0\)
\(\Delta=\left(-3\right)^2-4\left(-8\right).2=73\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{3-\sqrt{73}}{4}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{73}}{4}\end{matrix}\right.\)
\(TH_2:t=-9\Rightarrow2x^2-3x-9=-9\Rightarrow2x^2-3x=0\Rightarrow x\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)