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1) ( x - y)2 - ( x + y)2 = -4xy
\(\Leftrightarrow\)( x - y - x + y ) ( x - y + x + y ) = -4xy
\(\Leftrightarrow\)2x + 4xy = 0
\(\Leftrightarrow\)2x ( 1 + 2y ) = 0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=0\\1+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}0\\-\dfrac{1}{2}\end{matrix}\right.\)
2) ( 7n -2)2 - ( 2n - 7)2
= ( 7n - 2 - 2n - 7 )( 7n - 2 + 2n - 7 )
= ( 5n - 9 )( 9n - 9 )
Ta có: 9n \(⋮\) 9 với mọi n
9 \(⋮\) 9 với mọi n
\(\Rightarrow\)9n - 9 \(⋮\) 9 với mọi n
\(\Rightarrow\) đpcm
3) F = x2 + 6x + 1
F = x2 + 2.x.3 + 9 - 8
F = ( x + 3 )2 - 8
Vì ( x + 3)2 \(\ge\) 0 với mọi x
\(\Rightarrow\) ( x + 3 )2 - 8 \(\ge\) -8 với mọi x
\(\Rightarrow\) F \(\ge\) -8 với mọi x
Vậy min F = -8 \(\Leftrightarrow\) ( x + 3 )2 = 0
\(\Leftrightarrow\) x = -3
1. Ta có: \(\left(x-y\right)^2-\left(x+y\right)^2=\left(x-y+x+y\right)\left(x-y-x-y\right)=2x.\left(-2y\right)=-4xy\)
2. Ta có: \(\left(7n-2\right)^2-\left(2n-7\right)^2=\left(7n-2-2n+7\right)\left(7n-2+2n-7\right)=\left(5n+5\right)\left(9n-9\right)=9\left(n-1\right)\left(5n+5\right)\)
\(\Rightarrow\left(7n-2\right)^2-\left(2n-7\right)^2\) chia hết cho 9 với mọi giá trị nguyên của n.
3. Ta có: \(F=-x^2+6x+1=-\left(x^2-6x-1\right)=-\left(x^2-6x+9-10\right)=-\left(x-3\right)^2+10\)
Vì \(-\left(x-3\right)^2\le0\Rightarrow-\left(x-3\right)^2+10\le10\)
=> MaxF=10 <=> \(-\left(x-3\right)^2+10=10\Leftrightarrow-\left(x-3\right)^2=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy MaxF=10 khi x=3.
4. Ta có: \(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2-a^2x^2-2abxy-b^2y^2=0\Leftrightarrow a^2y^2+b^2x^2-2abxy=0\Leftrightarrow\left(ay-bx\right)^2=0\Leftrightarrow ay-bx=0\)
=> đpcm.
a) (x-2)^3-x(x+1)(x-1)+6x(x-3)=0
\(x^3-6x^2+12x-8-x\left(x^2-1\right)+6x\left(x-3\right)=0\)
\(x^3-6x^2+12x-8-x^3+x+6x^2-18x=0\)
\(-5x-8=0\)
\(x=-\frac{8}{5}\)
Mai mik làm mấy bài kia sau
Bài 1:
a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(\Leftrightarrow2x^2-10x-3x-2x^2-26=0\)
\(\Leftrightarrow-13x-26=0\)
\(\Leftrightarrow-13\left(x+2\right)=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
b) \(\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-7\end{matrix}\right.\)
Bài 2:
a) \(\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
b) \(\left(2x-1\right)\left(2x+1\right)\left(1-5x\right)\)
\(=\left(4x^2-1\right)\left(1-5x\right)\)
\(=4x^2-20x^3-1+5x\)
B1:
a) \(\left(10x+9\right)x-\left(5x-1\right)\left(2x+3\right)=8\)
\(10x^2+9x-10x^2-15x+2x+3-8=0\)
\(-4x-5=0\)
\(-4x=5\Leftrightarrow x=-\dfrac{5}{4}\)
b) \(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\)
\(21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(42x-41=0\)
\(x=\dfrac{41}{42}\)
3.
\(x=\left|2\right|\Rightarrow x=\pm2\)
Thay x = 2 vào A ta có:
A = (3.2+5)(2.2+1) + (4.2+1)(5.2+2)
= 11.5 + 9.12
= 55 + 108
= 163
Thay x = -2 vào A ta có:
A = (-2.3+5)(-2.2+1) + (-2.4+1)(-2.5+2)
= (-1)(-3) + (-7)(-8)
= 3 + 56
= 59
Thay x = -1 vào B ta có:
B = (-1-3)(-1+7) - (-1.2-5)(-1-1)
= (-4).6 - (-7)(-2)
= -24 - 14
= -38
Vậy \(A=163\Leftrightarrow x=2\)
\(A=59\Leftrightarrow x=-2\)
\(B=-38\Leftrightarrow x=-1\)
\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
a/ \(x^2+y^2=x^2+y^2+2xy-2xy =\left(x+y\right)^2-2xy\)
b/ mình không chắc nữa
bài 3
a/ \(9x^2-49=0 \Leftrightarrow x^2=\frac{49}{9} \Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{7}{3}\end{cases}}\)
b/ \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x+1\right)\left(x-1\right)-27=0 \Leftrightarrow x^3+27-x\left(x^2-1\right)-27=0\)
\(\Leftrightarrow x^3-x^3+x=0\Leftrightarrow x=0\)
c/\(\left(x-1\right)\left(x+2\right)-x-2=0 \Leftrightarrow \left(x-1\right)\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)^2=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
d/ \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\)
\(\Leftrightarrow4x+25=0 \Leftrightarrow x=\frac{-25}{4}\)
e/ mình lười qá ko viết đề đâu
\(\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\)
\(\Leftrightarrow-3x+1=7 \Leftrightarrow x=-2\)
có gì sai bn sửa lại nha
1: =>(x+3)(x-5)=0
=>x=5 hoặc x=-3
2: =>(x-1)(5x-1)=0
=>x=1/5 hoặc x=1
5: =>(x-4)*x=0
=>x=0 hoặc x=4
10: =>(x+5)(x-3)=0
=>x=3 hoặc x=-5
9: =>(x-2)(x-4)=0
=>x=2 hoặc x=4
7: =>(x-6)(2x-1)=0
=>x=1/2 hoặc x=6
8: =>(2x-1)(3x-12)=0
=>x=4 hoặc x=1/2
\(A=x^2-6x+10\)
\(=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\)
\(\left(x-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+1\ge1>0\)
Vậy A > 0 với mọi x.
\(B=x^2-2xy+y^2+1\)
\(=\left(x-y\right)^2+1\)
\(\left(x-y\right)^2\ge0\)
\(\Rightarrow\left(x-y\right)^2+1\ge1>0\)
Vậy B > 0 với mọi x, y.
\(M=x^2-6x+12\)
\(=x^2-6x+9+3\)
\(=\left(x-3\right)^2+3\)
\(\left(x-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+3\ge3\)
\(MinB=3\Leftrightarrow x=3\)
\(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)
\(x^2+6x+9+x^2-4-2\left(x^2-2x+1\right)=7\)
\(2x^2+6x+5-2x^2+4x-2=7\)
\(10x=7+3\)
\(10x=10\)
\(x=1\)
\(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x+1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)
\(x^3-\frac{1}{4}x=0\)
\(x\left(x^2-\frac{1}{4}\right)=0\)
\(x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{array}\right.\)
\(\left(x+10\right)^2-\left(x^2+2x\right)\)
\(=x^2+20x+100-x^2-2x\)
\(=18x+100\)
\(\left(x+2\right)\left(x-2\right)+\left(x-1\right)\left(x^2+x+1\right)-x\left(x^2+x\right)\)
\(=x^2-4+x^3-1-x^3-x^2\)
\(=-5\)
bài 2:
câu 2:
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Rightarrow\left(2x\right)^2-2\cdot2x\cdot1+1^2+x^2+2\cdot x\cdot3+3^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Rightarrow4x^2-4x+1+x^2+6x+9-5\left(x-7\right)\left(x+7\right)=0\)
\(\Rightarrow5x^2+2x+10-5\left(x^2+7x-7x-49\right)=0\)
\(\Rightarrow5x^2+2x+10-5\left(x^2-49\right)=0\)
\(\Rightarrow5x^2+2x+10-5x^2+245=0\)
\(\Rightarrow2x-255=0\)
\(\Rightarrow2x=255\Rightarrow x=255:2=\frac{255}{2}=127,5\)
ko chắc lắm!!!