a(x-y)2-x+y
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Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)
\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)
Dấu "=" \(x=y=\dfrac{1}{2}\)
Trả lời:
7, 5( x + y )2 + 15( x + y )
= 5( x + y )( x + y + 3 )
9, 7x( y - 4 )2 - ( 4 - y )3
= 7x ( 4 - y )2 - ( 4 - y )
= ( 4 - y )2 ( 7x - 4 + y )
11, ( x + 1 )( y - 2 ) - ( 2 - y )2
= ( x + 1 )( y - 2 ) - ( y - 2 )2
= ( y - 2 )( x + 1 - y + 2 )
= ( y - 2 )( x - y + 3 )
8, 9x ( x - y ) - 10 ( y - x )2
= 9x ( x - y ) - 10 ( x - y )2
= ( x - y )[ ( 9x - 10 ( x - y ) ]
= ( x - y )( 9x - 10x + 10y )
= ( x - y )( 10y - x )
10, ( a - b )2 - ( a + b )( b - a )
= ( b - a )2 - ( a + b )( b - a )
= ( b - a )( b - a - a - b )
= - 2a( b - a )
= 2a ( a - b )
12, 2x ( x - 3 ) + y ( x - 3 ) + ( 3 - x )
= 2x ( x - 3 ) + y ( x - 3 ) - ( x - 3 )
= ( x - 3 )( 2x + y - 1 )
Câu 16: Chọn câu sai.
A. (x + y)2 = (x + y)(x + y)
B. x2 – y2 = (x + y)(x – y)
C. (-x – y)2 = (-x)2 – 2(-x)y + y2
D. (x + y)(x + y) = y2 – x2
Câu 17: Chọn câu đúng
A. (c + d)2 – (a + b)2 = (c + d + a + b)(c + d – a + b)
B. (c – d)2 – (a + b)2 = (c – d + a + b)(c – d – a + b)
C. (a + b + c – d)(a + b – c + d) = (a + b)2 – (c – d)2
D. (c – d)2 – (a – b)2 = (c – d + a – b)(c – d – a – b)
Câu 18: Có bao nhiêu giá trị x thỏa mãn (2x – 1)2 – (5x – 5)2 = 0
A. 0 B. 1 C. 2 D. 3
Câu 19: Có bao nhiêu giá trị x thỏa mãn (2x + 1)2 – 4(x + 3)2 = 0
A. 0 B. 1 C. 2 D. 3
Câu 20:Tìm x biết (x – 6)(x + 6) – (x + 3)2 = 9
A. x = -9 B. x = 9 C. x = 1 D. x = -6
Câu 8: B
a: \(A=y^2-8y-x\left(8-y\right)\)
\(=y\left(y-8\right)+x\left(y-8\right)\)
\(=\left(y-8\right)\left(x+y\right)\)
\(=100\cdot100=10000\)
\(\left(a\right):\left(x+y\right)^2-\left(x-y\right)^2=x^2+2xy+y^2-\left(x^2-2xy+y^2\right)\\ =x^2+2xy+y^2-x^2+2xy-y^2\\ =4xy\)
\(\left(b\right):\left(x-y-z\right)^2+\left(x+y+z\right)^2\\ =\left[\left(x-y\right)-z\right]^2+\left[\left(x+y\right)+z\right]^2\\ =\left(x-y\right)^2-2z\left(x-y\right)+z^2+\left(x+y\right)^2+2z\left(x+y\right)+z^2\\ =x^2-2xy+y^2-2xz+2yz+z^2+x^2+2xy+y^2+2xz+2yz+z^2\\ =2x^2+2y^2+2z^2+4yz\)
\(\left(c\right):\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =\left(2y\right)^2=4y^2\)
a: =2(x-y)^3/(x-y)-7(x-y)^2/(x-y)+(x-y)/(x-y)
=2(x-y)^2-7(x-y)+1
b: =3(x-y)^5/5(x-y)^2-2(x-y)^4/5(x-y)^2+3(x-y)^2/5(x-y)^2
=3/5(x-y)^3-2/5(x-y)^2+3/5
\(a,\)
\(\left[2\left(x-y\right)^3-7\left(y-x\right)^2-\left(y-x\right)\right]:\left(x-y\right)\)
\(=\left[2\left(x-y\right)^3-7\left(x-y\right)^2+\left(x-y\right)\right]:\left(x-y\right)\)
\(=\left\{\left(x-y\right)\left[2\left(x-y\right)^2-7\left(x-y\right)+1\right]\right\}:\left(x-y\right)\)
\(=2\left(x-y\right)^2-7\left(x-y\right)+1\)
\(b,\)
\(\left[3\left(x-y\right)^5-2\left(x-y\right)^4+3\left(x-y\right)^2\right]:\left[5\left(x-y\right)^2\right]\)
\(=\dfrac{3}{5}\left(x-y\right)^3-\dfrac{2}{5}\left(x-y\right)^2+\dfrac{3}{5}\)
a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=\left(x-y\right)\left(x-y+2\right)+37\)(1)
Thay x-y=7 vào biểu thức (1), ta được:
\(A=7\cdot\left(7+2\right)+37=7\cdot9+37=100\)
Vậy: Khi x-y=7 thì A=100
b) Ta có: \(x+y=2\)
\(\Leftrightarrow\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+y^2+2xy=4\)
\(\Leftrightarrow2xy+10=4\)
\(\Leftrightarrow2xy=-6\)
\(\Leftrightarrow xy=-3\)
Ta có: \(A=x^3+y^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\)(2)
Thay x+y=2; \(x^2+y^2=10\) và xy=-3 vào biểu thức (2), ta được:
\(A=2\cdot\left(10+3\right)=2\cdot13=26\)
Vậy: Khi x+y=2 và \(x^2+y^2=10\) thì A=26
\(\Rightarrow A=x^2+2x+y^2-2y-2xy+37=x^2-2xy+y^2+2\left(x-y\right)+37=\left(x-y\right)^2+2\left(x-y\right)+37=7^2+2\cdot7+37=100\)
\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[x^2+y^2-\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}\right]=2\cdot\left[10+3\right]=2\cdot13=26\) \(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{x+z}{x}\right)=-\dfrac{z}{y}\cdot\dfrac{-x}{z}\cdot-\dfrac{y}{x}=-1\)