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Bài 9 : Tìm x, biết :
a, (x - 2)(x - 3) + (x - 2) - 1 = 0
\(\Leftrightarrow\left(x-2\right)\left(x-3+1\right)-1=0\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x-2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy x ={1; 3}
b, (x + 2)2 - 2x(2x + 3) = (x + 1)2
\(\Leftrightarrow\left(x+2\right)^2-\left(x+1\right)^2-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(x+2+x+1\right)\left(x+2-x-1\right)-2x\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3-2x\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(1-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\1-2x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{-\frac{3}{2};\frac{1}{2}\right\}\)
c, 6x3 + x2 = 2x
\(\Leftrightarrow6x^3+x^2-2x=0\)
\(\Leftrightarrow x\left(6x^2+x-2\right)=0\)
\(\Leftrightarrow x\left(6x^2+4x-3x-2\right)=0\)
\(\Leftrightarrow x\left[2x\left(3x+2\right)-\left(3x+2\right)\right]=0\)
\(\Leftrightarrow x\left(3x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3x+2=0\\2x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{2}{3}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(x=\left\{0;-\frac{2}{3};\frac{1}{2}\right\}\)
a/ \(x^2\left(x-5\right)+5-x=0\)
\(\Leftrightarrow x^2\left(x-5\right)-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
Vậy...
b/ \(3x^4-9x^3=-9x^2+27x\)
\(\Leftrightarrow3x^4-9x^3+9x^2-27x=0\)
\(\Leftrightarrow3x^3\left(x-3\right)+9x\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x^3+9x\right)=0\)
\(\Leftrightarrow3x\left(x-3\right)\left(x^2+3\right)=0\)
Vì \(x^2+3>0\forall x\)
\(\Leftrightarrow3x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy..
c/ \(x^2\left(x+8\right)+x^2=-8x\)
\(\Leftrightarrow x^2\left(x+8\right)+x^2+8x=0\)
\(\Leftrightarrow x^2\left(x+8\right)+x\left(x+8\right)=0\)
\(\Leftrightarrow x\left(x+8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+8=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\\x=-1\end{matrix}\right.\)
Vậy...
d/ \(\left(x+3\right)\left(x^2-3x+5\right)=x^2+3x\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[\left(x-2\right)^2+1\right]=0\)
Vì \(\left(x-2\right)^2+1>0\forall x\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy..
Úi, câu d bạn nên làm theo cách của bạn trên đúng hơn nha :< Mình nghĩ câu d mình lập luận bị sai rồi ó
Sửa lại ạ!
a) \(\left(3x-1\right)^2-16\)
\(=\left(3x-1\right)^2-4^2\)
\(=\left(3x-1-4\right)\left(3x-1+4\right)\)
\(=\left(3x-5\right)\left(3x+3\right)\)
b) \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4\right)^2-\left(7x\right)^2\)
\(=\left(5x-4-7x\right)\left(5x-4+7x\right)\)
\(=\left(-4-2x\right)\left(-4+12x\right)\)
c) \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+14\right)\left(3x-4\right)\)
d) \(\left(3x+1\right)^2-4\left(x-2\right)^2\)
\(=\left(3x+1\right)^2-\left[2\left(x-2\right)\right]^2\)
\(=\left(3x+1\right)^2-\left(2x-4\right)^2\)
\(=\left(3x+1-2x+4\right)\left(3x+1+2x-4\right)\)
\(=\left(x+5\right)\left(5x-3\right)\)
e) \(9\left(2x+3\right)^2-4\left(x+1\right)^2\)
\(=\left[3\left(2x+3\right)\right]^2-\left[2\left(x+1\right)\right]^2\)
\(=\left(6x+9\right)^2-\left(2x+2\right)^2\)
\(=\left(6x+9-2x-2\right)\left(6x+9+2x+2\right)\)
\(=\left(4x+7\right)\left(8x+11\right)\)
P/s: Ko chắc!
Bài 4 : Tính nhanh :
a, 15. 64 + 25. 100 + 36. 15 + 60. 100
= (15 . 64 + 36. 15) + (25. 100 + 60. 100)
= 15.(64 + 36) + 100.(25 + 60)
= 15. 100 + 100. 85
= 100.(15 + 85)
= 100. 100
= 10000
b, 472 + 482 - 25 + 94. 48
= 472 + 2.47. 48 + 482 - 25
= (47 + 48)2 - 52
= (47 + 48 - 5)(47 + 48 + 5)
= (48 + 22)(48 + 52)
= 90. 100
= 9000
c, 93 - 92. ( -1) - 9. 11 + ( -1). 11
= 93 + 92 + 11(- 9 - 1)
= 92.(9 + 1) + 11. (-10)
= 81. 10 - 110
= 810 - 110
= 700
d,2016. 2018 - 20172
= (2017 - 1)(2017 + 1) - 20172
= 20172 - 1 - 20172
= -1
#Học tốt!
Bài 4:
a) Ta có: \(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)
\(=\left(x^9-x^7\right)-\left(x^6-x^4\right)-\left(x^5-x^3\right)+\left(x^2-1\right)\)
\(=x^7\left(x^2-1\right)-x^4\left(x^2-1\right)-x^3\left(x^2-1\right)+\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^7-x^4-x^3+1\right)\)
\(=\left(x^2-1\right)\cdot\left[x^4\left(x^3-1\right)-\left(x^3-1\right)\right]\)
\(=\left(x^2-1\right)\cdot\left(x^3-1\right)\cdot\left(x^4-1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x-1\right)\left(x^2+x+1\right)\cdot\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
\(=\left(x-1\right)^3\cdot\left(x+1\right)^2\cdot\left(x^2+1\right)\cdot\left(x^2+x+1\right)\)
a, Ta có : \(x^5-x^4-x^3-x^2-x-2\)
\(=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2\)
\(=x^4\left(x-2\right)+x^3\left(x-2\right)+x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)\)
\(=\left(x-2\right)\left(x^4+x^3+x^2+x+1\right)\)
a, - Đặt \(x^2+4x+8=a\) ta được :\(a^2+3xa+2x^2\)
\(=a^2+xa+2xa+2x^2\)
\(=a\left(a+x\right)+2x\left(a+x\right)\)
\(=\left(2x+a\right)\left(x+a\right)\)
- Thay lại x vào đa thức ta được :
\(\left(2x+x^2+4x+8\right)\left(x+x^2+4x+8\right)\)
\(=\left(x^2+6x+8\right)\left(x^2+5x+8\right)\)
b, - Đặt \(x^2+x+1=a\) ta được :\(a\left(a+1\right)-12\)
\(=a^2+a-12\)
\(=a^2+\frac{1}{2}.2.a+\frac{1}{4}-\frac{49}{4}\)
\(=\left(a+\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)
\(=\left(a+\frac{1}{2}+\frac{7}{2}\right)\left(a+\frac{1}{2}-\frac{7}{2}\right)\)
\(=\left(a+4\right)\left(a-3\right)\)
- Thay lại x vào đa thức ta được :
\(\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
c, - Đặt \(x^2+8x+7=a\) ta được : \(a\left(a+8\right)+15\)
\(=a^2+8a+15\)
\(=a^2+3a+5a+15\)
\(=a\left(a+3\right)+5\left(a+3\right)\)
\(=\left(a+3\right)\left(a+5\right)\)
- Thay lại x vào đa thức ta được :
\(\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
d, Ta có : \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+2x+5x+10\right)\left(x^2+3x+4x+12\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
- Đặt \(x^2+7x+10=a\) ta được : \(a\left(a+2\right)-24\)
\(=a^2+2a-24\)
\(=a^2-4a+6a-24\)
\(=a\left(a-4\right)+6\left(a-4\right)\)
\(=\left(a+6\right)\left(a-4\right)\)
- Thay lại x vào đa thức ta được :
\(\left(x^2+7x+10+6\right)\left(x^2+7x+10-4\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)
Bài 1 : Tìm x,biết :
a, x2(x + 5) - 9x = 45
⇔ x2(x + 5) - 9x - 45 = 0
⇔ x2(x + 5) - 9(x + 5) = 0
⇔ (x + 5)(x2 - 9) = 0
⇔ (x + 5)(x - 3)(x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\\x=-3\end{matrix}\right.\)
Vậy x ={-5; 3; -3}
b, 9(5 - x) + x2 - 10x = -25
⇔ 45 - 9x + x2 - 10x + 25 = 0
⇔ x2 - 19x + 70 = 0
⇔ x2 - 14x - 5x + 70 = 0
⇔ (x2 - 5x) - (14x - 70) = 0
⇔ x(x - 5) - 14(x - 5) = 0
⇔ (x - 5)(x - 14) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)
Vậy x ={5; 14}
a, x2( x+5 ) - 9x = 45
x3 + 5x2 - 9x - 45 = 0
x2( x+5 ) - 9( x+5) = 0
(x2 - 9)(x + 5) = 0
(x + 3)(x - 3)(x + 5) = 0
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-5\end{matrix}\right.\)
b, 9( 5-x ) + x2 -10x = -25
45 - 9x + x2 - 10x + 25 = 0
x2 - 19x + 70 = 0
x2 - 14x - 5x + 70 = 0
x( x-14 ) - 5( x-14) = 0
(x - 5)(x - 14) = 0
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)