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mình làm bài 2 trước nha:
a) y.(a-b)+a.(y-b)=a.y-b.y+a.y-b.y
=(a.y+a.y)-(b.y+b.y)
=2.a.y-2.b.y
=2.y.(a-b)
b)x2.(x+y)-y.(x2-y2)=x3+x2.y-x2y+y3=x3+y3
\(x=\frac{1}{2}\frac{\sqrt{\left(\sqrt{2}-1\right)^2}}{\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}}=\frac{1}{2}.\left(\sqrt{2}-1\right)\)
\(\Rightarrow2x=\sqrt{2}-1\Rightarrow2x+1=\sqrt{2}\)
\(\Rightarrow4x^2+4x+1=2\Rightarrow4x^2+4x-1=0\)
\(B=\left[x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+4x^2+4x-1-1\right]^{2018}+2018\)
\(=\left(-1\right)^{2018}+2018=2019\)
a) \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x\left(x^2-5x+1\right)-2\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-5x^2+x-2x^2+10x-2-x^3-11x\)
\(=-7x^2-2\)
b) \(\left(x-1\right)\left(x^2+x+1\right)+x^3-2\)
\(=x\left(x^2+x+1\right)-1\left(x^2+x+1\right)+x^3-2\)
\(=x^3+x^2+x-x^2-x-1+x^3-2\)
\(=2x^3-3\)
c) \(\left(x-y\right)\left(x+y\right)-2x\left(x-y\right)\)
\(=x\left(x+y\right)-y\left(x+y\right)-2x\left(x-y\right)\)
\(=x^2+xy-yx-y^2-2x^2+2xy\)
\(=-x^2-y^2+2xy\)
a, \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-7x^2+11x-2-x^3-11x=-7x^2-2\)
b, \(\left(x-1\right)\left(x^2+x+1\right)+\left(x^3-2\right)\)
\(=x^3-1+x^3-2=2x^3-3\)
c, \(\left(x-y\right)\left(x+y\right)-2x\left(x-y\right)\)
\(=x^2-y^2-2x^2+2xy=-x^2-y^2+2xy\)
a, -5x^2 - x + 6 = 0
<=> -5x^2 - x -5 -1 =0
<=> (-5x^2-5)-(x+1)=0
<=> -5(x^2-1)-(x+1)=0
<=> -5(x-1)(x+1)-(x+1)=0
<=> (x+1)(-5x+5-1)=0
<=> (x+1)(-5x+4)=o
<=> x+1=0 hoặc -5x+4=0
*) x+1=0<=>x=-1
*) -5x+4=0<=> x=4/5
b) 4x^2+x-5= 0
<=> 4x^2 - 4 +x-1=0
<=> 4(x^2-1)+(x-1)=0
<=> 4(x-1)(x+1)+(x-1)=0
<=>(x-1)(4x+4+1)=0
<=>(x-1)(4x+5)=0
<=> x-1=0 hoặc 4x+5=0
+) x-1=0<=> x=1
+)4x+5=0<=>x=-5/4
...cách làm .. ptđt thành nhân tử
.giải .... >>
a . \(-5x^2-x+6=0\)
<=> (-5x^2 +5x ) - ( 6x-6)=0 <=> -5x(x-5) -6 (x-1) =0 <=> (x-1)(-5x-6/5)=0 <=> x=1 hoặc x =-6/5
b.<=> 4x^2-4x +5x-5 = 0 <=.> 4x(x-1) +5 (x-1) <=> (x-1)(4x+5)=0 <=> x= 1 hoặc x= -5/4
...... chuẩn ko cần chỉnh .. check liền tay ..The end•••
a) \(x\left(2x+1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)=3\)
\(\Leftrightarrow2x^2+x-x^3-2x^2+x^3-x+3=3\)
\(\Leftrightarrow3=3\)( Luôn đúng với mọi x )
Vậy phương trình nghiệm đúng với mọi x
b) \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x\left(x-1\right)=12x+12\)
\(\Leftrightarrow4x-24-2x^2-3x^3+5x^2-4x+3x^2-3x=12x+12\)
\(\Leftrightarrow-3x^3+6x^2-3x-24=12x+12\)
\(\Leftrightarrow-3x^3+6x^2-3x-24-12x-12=0\)
\(\Leftrightarrow-3x^3+6x^2-15x-36=0\)
Đến đây xem lại đề bạn nhớ :D Tìm thì tìm được nhưng thấy nó sai sai kiểu gì í
c) \(\left(3x+1\right)\left(x-2\right)=\left(2-x\right)\left(-3x-5\right)\)
\(\Leftrightarrow3x\left(x-2\right)+1\left(x-2\right)=2\left(-3x-5\right)-x\left(-3x-5\right)\)
\(\Leftrightarrow3x^2-6x+x-2=-6x-10+3x^2+5x\)
\(\Leftrightarrow3x^2-6x+x+6x-3x^2-5x=-10+2\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\)
d) \(\left(x+3\right)\left(x+5\right)-x\left(x+7\right)=2x+8\)
\(\Leftrightarrow x\left(x+5\right)+3\left(x+5\right)-x\left(x+7\right)=2x+8\)
\(\Leftrightarrow x^2+5x+3x+15-x^2-7x=2x+8\)
\(\Leftrightarrow x^2+5x+3x-x^2-7x-2x=8-15\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
a, \(x\left(2x-1\right)-x^2\left(x+2\right)+\left(x^3-x+3\right)=3\)
\(\Leftrightarrow2x^2-x-x^3-2x^2+x^3-x+3=3\)
\(\Leftrightarrow-2x=0\Leftrightarrow x=0\)
b, \(4\left(x-6\right)-x^2\left(2+3x\right)+x\left(5x-4\right)+3x\left(x-1\right)=12x+12\)
\(\Leftrightarrow4x-24-2x^2-3x^3+5x^2-4x+3x^2-3x=12x+12\)
\(\Leftrightarrow-3x-24+6x^2-3x^3=12x+12\)
\(\Leftrightarrow-15x-36+6x^2-3x^3=0\)
Lớp 8 chưa hc vô tỉ đâu ... vô nghiệm
c, \(\left(3x+1\right)\left(x-2\right)=\left(2-x\right)\left(-3x-5\right)\)
\(\Leftrightarrow3x^2-5x-2=-x-10+3x^2\)
\(\Leftrightarrow-4x+8=0\Leftrightarrow x=2\)
d, \(\left(x+3\right)\left(x+5\right)-x\left(x+7\right)=2x+8\)
\(\Leftrightarrow x^2+8x+15-x^2-7x=2x+8\)
\(\Leftrightarrow x+15=2x+8\Leftrightarrow-x+7=0\Leftrightarrow x=7\)
Ta có: \(\frac{a+1}{b^2+1}=\left(a+1\right)-\frac{\left(a+1\right)b^2}{b^2+1}\)
\(\ge\left(a+1\right)-\frac{\left(a+1\right)b^2}{2b}=a+1-\frac{ab+b}{2}\)
Tương tự ta có:\(\frac{b+1}{c^2+1}\ge b+1-\frac{bc+c}{2};\frac{c+1}{a^2+1}\ge c+1-\frac{ca+a}{2}\)
Cộng theo vế ta có: \(VT\ge a+b+c+3-\frac{ab+bc+ca+a+b+c}{2}=6-\frac{3+ab+bc+ca}{2}\)
Mà theo BĐT AM-GM: \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)
Suy ra \(VT\ge6-3=3\)(ĐPCM)
a)\(\left(-x^2y^5\right)^2:\left(-x^2y^5\right)=\left(-x^2y^5\right)\)
b)\(5\cdot\left(x-2y\right)^3:\left(5x-10y\right)\)
\(=5\cdot\left(x-2y\right)\cdot\left(x-2y\right)^2:\left(5x-10y\right)\)
\(=\left(5x-10y\right)\cdot\left(x-2y\right)^2:\left(5x-10y\right)\)
\(=\left(x-2y\right)^2\)
Thay \(x=\frac{1}{2},y=1\) vào:
\(\left(\frac{1}{2}-2\cdot1\right)^2=\left(\frac{-3}{2}\right)^2=\frac{9}{4}\)