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a,
\(4\left(6-x\right)+x^2\left(2+3x\right)-x\left(5x-4\right)+3x^2\left(1-x\right)\)
= \(24-4x+2x^2+3x^3-5x^2-4x+3x^2-3x^3\)
\(=24-\left(4x-4x\right)+\left(2x^2-5x^2+3x^2\right)+\left(3x^3-3x^3\right)\)
\(=24\)
vì kết quả không phụ thuộc vào giá trị của biến
=> (đpcm)
a) https://hoc24.vn/hoi-dap/question/398481.html
b)
a2 + b2 + c2 = ab + ac + bc
<=> 2a2 + 2b2 + 2c2 = 2ac + 2ab + 2bc
<=> (a2 - 2ac + c2) + (a2 - 2ab + b2) + (b2 - 2bc + c2) = 0
<=> (a - b)2 + (a - c)2 + (b - c)2 = 0
<=> a = b = c
1. Ta có:
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
=> \(a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)
=> \(a^2y^2+b^2x^2=2axby\)
=> \(a^2y^2+b^2x^2-2axby=0\)
=> \(a^2y^2+b^2x^2-2aybx=0\)
=> \(\left(ay-bx\right)^2=0\)
Mà \(\left(ay-bx\right)^2\ge0\)
Dấu '' = '' xảy ra \(\Leftrightarrow\) \(ay-bx=0\)
\(\Leftrightarrow\) \(ay=bx\)
\(\Leftrightarrow\) \(\dfrac{a}{x}=\dfrac{b}{y}\)
2. Ta có:
\(a^2+b^2+c^2=ab+bc+ac\)
=> \(2a^2+2b^2+2c^2=2ab+2bc+2ac\)
=> \(2a^2+2b^2+2c^2-2ab-2bc-2ac=0\)
=> \(\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+\left(b^2-2bc+c^2\right)=0\)
=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
Ta thấy:
\(\left(a-b\right)^2\ge0\); \(\left(a-c\right)^2\ge0\); \(\left(b-c\right)^2\ge0\)
=> \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\)
Mà \(\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2=0\)
Dấu '' = '' xảy ra \(\Leftrightarrow\) \(\left\{{}\begin{matrix}a-b=0\\a-c=0\\b-c=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=b\\a=c\\b=c\end{matrix}\right.\)
\(\Leftrightarrow\) a = b = c
\(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
Thay x - y = 7
\(\Rightarrow A=49+14+37=100\)
Vậy A = 100 khi x - y = 7
a) A = x2( x + y ) - y( x2 + y2 )
= x3 + x2y - x2y - y3
= x3 - y3
Với x = 1 ; y = -1
A = 13 - (-1)3 = 1 + 1 = 2
b) B = 5x( x - 4y ) - 4y( y - 5x )
= 5x2 - 20xy - 4y2 + 20xy
= 5x2 - 4y2
Với x = -0, 6 ; y = -0, 75
B = 5.(-0, 6)2 - 4.(-0, 75)2 = 5.9/25 - 4.9/16 = 9/5 - 9/4 = -9/20
C = x( x - y + 1 ) - y( y + 1 - x )
= x2 - xy + x - y2 - y + xy
= x2 + x - y2 - y
= ( x2 - y2 ) + ( x - y )
= ( x - y )( x + y ) + ( x - y )
= ( x - y )( x + y + 1 )
Thế x = -2/3 ; y = -1/3 ta được
C = [ -2/3 - (-1/3 ) ][ -2/3 - 1/3 + 1 ]
= ( -2/3 + 1/3 ).0
= 0
a, \(A=x^2\left(x+y\right)-y\left(x^2+y^2\right)+2002=x^3-y^3+2002\)
Thay x = 1; y = -1 ta có : \(1^3-\left(-1\right)^3+2002=1-1+2002=2002\)
b, \(5x\left(x-4y\right)-4y\left(y-5x\right)-\frac{11}{20}=5x^2-4y^2-\frac{11}{20}\)
Thay x = -0,6 ; y = -0,75 ta có : \(5.\left(-0,6\right)^2-4\left(-0,75\right)^2-\frac{11}{20}=-1\)
c, \(x\left(x-y+1\right)-y\left(y+1-x\right)=x^2+x-y^2-y\)
Thay x = -2/3 ; y = -1/3 ta có : \(\left(-\frac{2}{3}\right)^2-\frac{2}{3}-\left(-\frac{1}{3}\right)^2+\frac{1}{3}=0\)
a, x.(x-y) +y.(x+y)
=x2-xy+xy+y2
=x2+y2
b, (x2-5).(2x+3)-2x.(x-3)
=2x3+3x2-10x-15-2x2+6x
=2x3-x2-4x-15
c, 8-5x.(x+2) +4 .( x-2) . (x+1) +2.( x+2)+ 2.(x-2)+10
=8-5x2-10x+4.(x2+x-2x-2)+2x+4+2x-4+10
=18-6x-5x2+4x2+4x-8x-8
=10-10x-x2
a) \(\left(3x-1\right)\left(2x+7\right)-\left(x+1\right)\left(6x-5\right)=16\) (1)
\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2-5x+6x-5\right)=16\)
\(\Leftrightarrow6x^2+21x-2x-7-\left(6x^2+x-5\right)=16\)
\(\Leftrightarrow6x^2+21x-2x-7-6x^2-x+5=16\)
\(\Leftrightarrow18x-2=16\)
\(\Leftrightarrow18x=16+2\)
\(\Leftrightarrow18x=18\)
\(\Leftrightarrow x=1\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{1\right\}\)
b) \(\left(10x+9\right)\cdot x-\left(5x-1\right)\left(2x+3\right)=8\) (2)
\(\Leftrightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)
\(\Leftrightarrow10x^2+9x-\left(10x^2+13x-3\right)=8\)
\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)
\(\Leftrightarrow-4x+3=8\)
\(\Leftrightarrow-4x=8-3\)
\(\Leftrightarrow-4x=5\)
\(\Leftrightarrow x=-\dfrac{5}{4}\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{-\dfrac{5}{4}\right\}\)
c) \(\left(3x-5\right)\left(7-5x\right)+\left(5x+2\right)\left(3x-2\right)-2=0\) (3)
\(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
\(\Leftrightarrow42x-41=0\)
\(\Leftrightarrow42x=41\)
\(\Leftrightarrow x=\dfrac{41}{42}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{41}{42}\right\}\)
d) \(x\left(x+1\right)\left(x+6\right)-x^3=5x\) (4)
\(\Leftrightarrow\left(x^2+x\right)\left(x+6\right)-x^3=5x\)
\(\Leftrightarrow x^3+6x^2+x^2+6x-x^3=5x\)
\(\Leftrightarrow7x^2+6x=5x\)
\(\Leftrightarrow7x^2+6x-5x=0\)
\(\Leftrightarrow7x^2+x=0\)
\(\Leftrightarrow x\left(7x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\7x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{7}\end{matrix}\right.\)
Vậy tập nghiệm phương trình (4) là \(S=\left\{-\dfrac{1}{7};0\right\}\)
a) Thay trực tiếp x=1, y= -1 vào cho nhanh :D
Khi đó A = 12. (1-1) - (-1). [12+(-1)2]
A = 1.0+1.2 = 2
b) B=5x .(x-4y)-4y .(y-5x)- \(\dfrac{11}{20}\)
B = 5x2 - 4y2 - \(\dfrac{11}{20}\)
Thay x = -0,6, y = -0,75 ta đc:
B = 5. (-0,6)2 - 4.(-0,75)2 - \(\dfrac{11}{20}\)
B = -1
c) C= x .(x-y+1)-y .(y+1-x)
C = x(x+1) - y(y+1)
Thay x= \(\dfrac{2}{3}\) , y=\(\dfrac{-1}{3}\) vào ta đc:
C = \(\dfrac{2}{3}\left(\dfrac{2}{3}+1\right)-\left(\dfrac{-1}{3}\right)\left(\dfrac{-1}{3}+1\right)\)
C = \(\dfrac{4}{3}\)
bạn có thể giải rõ câu a ra hộ mk đc ko