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2 tháng 10 2020
Sorry mình nhầm câu a
a) (2x - 1)2 + (x + 3)2 - 5(x + 7)(x - 7) = 0
b) (x + 2)(x2 - 2x + 4) - x(x2 + 2) = 15
c) (x + 3)3 - x(3x + 1)2 + (2x - 1)(4x2 - 2x + 1) = 28
d) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
Giải:
a) (2x - 1)2 + (x + 3)2 - 5(x + 7)(x - 7) = 0
\(\Leftrightarrow\) 4x2 - 4x + 1 + x2 + 6x + 9 - 5(x2 - 49) = 0
\(\Leftrightarrow\) 4x2 - 4x + 1 + x2 + 6x + 9 - 5x2 + 245 = 0
\(\Leftrightarrow\) 2x + 255 = 0
\(\Leftrightarrow\) 2x = - 255
\(\Leftrightarrow\) x = - 255 : 2
\(\Leftrightarrow\) x = \(-\frac{255}{2}\)
Vậy x = \(-\frac{255}{2}\)
b) (x + 2)(x2 - 2x + 4) - x(x2 + 2) = 15
\(\Leftrightarrow\) x3 + 8 - x3 - 2x = 15
\(\Leftrightarrow\) 8 - 2x = 15
\(\Leftrightarrow\) 2x = 8 - 1
\(\Leftrightarrow\) 2x = - 7
\(\Leftrightarrow\) x = - 7 : 2
\(\Leftrightarrow\) x = \(-\frac{7}{2}\)
Vậy x = \(-\frac{7}{2}\)
c) (x + 3)3 - x(3x + 1)2 + (2x - 1)(4x2 - 2x + 1) = 28
\(\Leftrightarrow\) x3 + 6x2 + 27x + 27 - x(9x2 + 6x + 1) + 8x3 - 1 = 28
\(\Leftrightarrow\) x3 + 6x2 + 27x + 27 - 9x3 - 6x2 - x + 8x3 - 1 = 28
\(\Leftrightarrow\) 26x + 26 = 28
\(\Leftrightarrow\) 26x = 28 - 26
\(\Leftrightarrow\) 26x = 2
\(\Leftrightarrow\) x = 2 : 26
\(\Leftrightarrow\) x = \(\frac{1}{13}\)
Vậy x = \(\frac{1}{13}\)
d) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
\(\Leftrightarrow\) x6 - 2x2 + 1 - (x6 - 1) = 0
\(\Leftrightarrow\) x6 - 2x2 + 1 - x6 + 1 = 0
\(\Leftrightarrow\) -2x2 + 2 = 0
\(\Leftrightarrow\) -2x2 = - 2
\(\Leftrightarrow\) x2 = - 2 : (- 2)
\(\Leftrightarrow\) x2 = 1
\(\Leftrightarrow\) x = 1 hoặc x = - 1
Vậy x \(\in\) {1; - 1}
16 tháng 11 2021
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
11 tháng 1 2021
a) Ta có: \(7x^2-28=0\)
\(\Leftrightarrow7\left(x^2-4\right)=0\)
\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)
mà 7>0
nên (x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)
mà \(\dfrac{2}{3}>0\)
nên x(x-2)(x+2)=0
hay \(\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;-2;2\right\}\)
c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)
\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)
d) Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-2\right\}\)
CM
4 tháng 12 2018
Bài 1:
a) \(x^2+9y^2-y^4-6xy\)
\(=\left(x^2-6xy+9y^2\right)-y^4\)
\(=\left[x^2-2.x.3y+\left(3y\right)^2\right]-\left(y^2\right)^2\)
\(=\left(x-3y\right)^2-\left(y^2\right)^2\)
\(=\left(x-3y-y^2\right)\left(x-3y+y^2\right)\)
b) \(2x^2-x-28\)
\(=2x^2-8x+7x-28\)
\(=2x\left(x-4\right)+7\left(x-4\right)\)
\(=\left(x-4\right)\left(2x+7\right)\)
Bài 2:
a) \(2x\left(x^2-2x+3\right)-2x^3\)
\(=2x\left(x^2-2x+3-x^2\right)\)
\(=2x\left(3-2x\right)\)
b) \(2x\left(x-3\right)-\left(x+5\right)\left(2x-1\right)\)
\(=\left(2x^2-6x\right)-\left(2x^2+9x-5\right)\)
\(=2x^2-6x-2x^2-9x+5\)
\(=-15x+5\)
\(=-5\left(3x-1\right)\)
c) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)
\(=\left(x-5\right)^2-2\left(x+5\right)\left(x-5\right)+\left(x+5\right)^2\)
\(=\left[\left(x-5\right)-\left(x+5\right)\right]^2\)
\(=\left(x-5-x-5\right)^2\)
\(=\left(-10\right)^2=100\)
Bài 3:
a) \(x-2=\left(x-2\right)^2\)
\(\Rightarrow\left(x-2\right)-\left(x-2\right)^2=0\)
\(\Rightarrow\left(x-2\right)\left(1-x+2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3-x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\left(-3x+9\right)x^2-7x+21=0\)
\(\Rightarrow-3\left(x-3\right)x^2-7\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(-3x^2-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\-3x^2-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x^2=-\dfrac{7}{3}\end{matrix}\right.\)
Mà x2 > 0 hoặc x2 = 0 với mọi x
=> x2 = -7/3 không thỏa mãn
=> x= 3
4 tháng 12 2018
Phân tích đa thức
a, x^2+9y^2-y^4-6xy
=(x^2-6xy+9y^2)-y^4
=(x-3y)^2-y^4
=(x-3y-y^2)(x-3y+y^2)
b, 2x^2-x-28
=(2x^2-8x)+(7x-28)
=2x(x-4)+7(x-4)
=(x-4)(2x+7)
Rút gọn
a,2x(x^2-2x+3)-2x^3
=2x(x^2-2x+3-x^2)
=2x(-2x+3)
b,2x(x-3)-(x+5)(2x-1)
=2x^2-6x-2x^2-9x+5
=-15x+5
=-5(3x-1)
c,(5-x)^2+(x+5)^2-(2x+10)(x-5)
Ta có:(5-x)^2=(x-5)^2
=(x-5)^2-2(x+5)(x-5)+(x+5)^2
=(x-5-x-5)^2
=100
Tìm x
a,x-2=(x-2)^2=0
=>x-2=0=>x=2
b,(-3x+9)x^2-7x+21=0
=>-3(x-3)x^2-7(x-3)=0
=>(x-3)(-3x^2-7)=0
=>\(\left[{}\begin{matrix}x-3=0=>x=3\\-3x^2-7=0=>x=\sqrt{\dfrac{-7}{3}}\end{matrix}\right.\)
18 tháng 6 2022
a: \(\Leftrightarrow\dfrac{x+5}{2x-1}+\dfrac{2x-1}{x+5}-2=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+5\right)+\left(2x-1\right)^2-2\left(2x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow x^2+10x+25+4x^2-4x+1-2\left(2x^2+10x-x-5\right)=0\)
\(\Leftrightarrow5x^2+6x+26-4x^2-18x+10=0\)
\(\Leftrightarrow x^2-12x+36=0\)
=>x=6
b: \(\dfrac{9x-27}{2x-7}-\dfrac{8x-28}{x-3}=0\)
\(\Leftrightarrow9\left(x-3\right)^2-4\left(2x-7\right)^2=0\)
\(\Leftrightarrow\left(3x-9\right)^2-\left(4x-14\right)^2=0\)
\(\Leftrightarrow\left(3x-9-4x+14\right)\left(3x-9+4x-14\right)=0\)
\(\Leftrightarrow\left(5-x\right)\left(7x-23\right)=0\)
hay \(x\in\left\{5;\dfrac{23}{7}\right\}\)
TD
1
9 tháng 12 2021
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
( 2x - 3 )2 - ( x + 5 )2 = 0
<=> [ ( 2x - 3 ) - ( x + 5 ) ][ ( 2x - 3 ) + ( x + 5 ) ] = 0
<=> ( 2x - 3 - x - 5 )( 2x - 3 + x + 5 ) = 0
<=> ( x - 8 )( 3x + 2 ) = 0
<=> \(\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
7x2 - 28 = 0
<=> 7( x2 - 4 ) = 0
<=> x2 - 4 = 0
<=> x2 = 4
<=> x2 = (±2)2
<=> x = ±2
a) \(\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left[\left(2x-3\right)-\left(x+5\right)\right].\left[\left(2x-3\right)+\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(2x-3-x-5\right).\left(2x-3+x+5\right)=0\)
\(\Leftrightarrow\left(x-8\right).\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-8=0\\3x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+8\\3x=0-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\3x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-\frac{2}{3}\end{cases}}\)
Vậy tập hợp nghiệm của phương trình là: \(S=\left\{8;-\frac{2}{3}\right\}.\)
b) \(7x^2-28=0\)
\(\Leftrightarrow7x^2=0+28\)
\(\Leftrightarrow7x^2=28\)
\(\Leftrightarrow x^2=28:7\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2-2^2=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0+2\\x=0-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
Vậy tập hợp nghiệm của phương trình là: \(S=\left\{2;-2\right\}.\)