|x + 9| + |x + 1| = 8 - (y^2 + 1)^2
Tính giá trị biểu thức:
a) x10+x9+x8+...+x tại x= -1
b)x100+x99+x98+...+x tại x= -1
c)x100-x99+x98+...+x2-x tại x=1
d)x10.y10+x9.y9+x8.y8+...+x.y tại x=1 và y= -1
e)x10.y10.z10+x9.y9.x9+x8.y8.z8+...+x.y.z tại x=-1, y= -1 và z=-1
f)3\(\sqrt{x-5}+7\) tại x=9
g)-5\(\sqrt{x^2-y^2}\) tại x=13 và y=12
h)4\(\sqrt{2.x^2+y^2-5}\) tại x=5 và y=6
a: \(=\left(-1\right)^{10}+\left(-1\right)^9+\left(-1\right)^8+...+\left(-1\right)^2+\left(-1\right)\)
\(=\left(1-1\right)+\left(1-1\right)+...+\left(1-1\right)\)
=0
b: \(=\left(-1\right)^{100}+\left(-1\right)^{99}+...+\left(-1\right)^2+\left(-1\right)\)
\(=\left(1-1\right)+...+\left(1-1\right)\)
=0
c: \(=1^{100}-1^{99}+1^{98}-1^{97}+...+1^2-1\)
=0
f: \(=3\cdot\sqrt{9-5}+7=3\cdot2+7=13\)
So sánh
a, 2011.2013+2012.2014 và 2012^2+2013^2-2
b, (9-1)(9^2+1)(9^4+1)(9^8+1)(9^16+1)(9^32+1) và 9^64-1
c, x-y/x+y và x^2-y^2/x^2+xy+y^2 với x>y>0
Giúp mình nha!!!
a) \(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
\(\Rightarrow2011.2013+2012.2014=2012^2+2013^2-2\)
b) \(\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9+1\right)\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^2-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^4-1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^8-1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{16}-1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{32}-1\right)\left(9^{32}+1\right)\)
\(=\dfrac{1}{10}\left(9^{64}-1\right)\)
\(=\dfrac{9^{64}-1}{10}\)
Ta có: \(9^{64}-1=\dfrac{10\left(9^{64}-1\right)}{10}\)
Mà \(\dfrac{10\left(9^{64}-1\right)}{10}>\dfrac{9^{64}-1}{10}\)
\(\Rightarrow\left(9-1\right)\left(9^2+1\right)\left(9^4+1\right)\left(9^8+1\right)\left(9^{16}+1\right)\left(9^{32}+1\right)< 9^{64}-1\)
c) Ta có:
\(\dfrac{x^2-y^2}{x^2+xy+y^2}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2-xy}\left(1\right)\)
Vì x>y>0, ta có:
\(\dfrac{x-y}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}\left(2\right)\)
Vì x>y>0 nên \(\left(x+y\right)^2-xy< \left(x+y\right)^2\left(3\right)\)
Từ (1), (2) và (3) suy ra:
\(\dfrac{x-y}{x+y}< \dfrac{x^2-y^2}{x^2+xy+y^2}\)
a) Ta có:
\(2011.2013+2012.2014\)
\(=\left(2012-1\right)\left(2012+1\right)+\left(2013-1\right)\left(2013+1\right)\)
\(=2012^2-1+2013^2-1\)
\(=2012^2+2013^2-2\)
Vậy 2011.2013+2012.2014 = 20122 + 20132 - 2
Bài 6:
a) A=y^2-8y-x(8-y) vs x=-8 y=108
b) B= y^2(x^2+y-1)- mx^2-my-m vs x=9 y= -80
c)C=x(y-x0^2-y(x-y)^2-y(x-y)^2+xy^2-x^2y vs x-y=7 xy=9
a: \(A=y^2-8y-x\left(8-y\right)\)
\(=y\left(y-8\right)+x\left(y-8\right)\)
\(=\left(y-8\right)\left(x+y\right)\)
\(=100\cdot100=10000\)
Bài 3: Rút gọn biểu thức (Dùng hằng đẳng thức)
1, (x+y)\(^2\)-(x-y)\(^2\)
2, (x+y)\(^3\)-(x-y)\(^3\)-2y\(^3\)
3,(x+y)\(^2\)-2(x+y)(x-y)+(x-y)\(^2\)
4,(2x+3)\(^2\)-2(2x+3)(2x+5)+(2x+5)\(^2\)
5, 9\(^8\). 2\(^8\)-(18\(^4\)+1)(18\(^4\)-1)
\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
Rút gọn 1) (x-2)^2 - (x+1) (x-1)
2)(3x-2) (3x+2)-(3x+1)^2
3)(x-8) (x+8)-(x+3)^2
4) (2x+y)^2 -2(4x^2-y^2)+(2x-y)^2
Bài 2 tìm x
a) (x+3)^2 - (x-2) (x+2)=8
b) (x-5)^2 - 9 =0
Bài 3 so sánh
A=(2+1) (2^2+1) (2^4+1) (2^8+1) và B = 2^16 .
Tìm x
a) x-6/7 + x-7/8 + x-8/9 = x-9/10 + x-10/11+x-11/12
b) x+32/11 + x+23/12 = x+38/13 + x+27/14
c) | x-2| = 13
d) 3|x-2| + |4x-8| = |-2| - |1/3|
e) |3x-2|+5^-1 = 3 + | x- (2/3) |
f) | x+2 | + | x-2 | = 3
g) (2x-1)^2 - 5 = 20
i) (x+2)^2 = 1/2 - 1/3
k) (x-1)^3 = (x-1)
m) (x-1)^x+2 = (x-1)^2
n) (x+3)^y+1 = (2x-1)^y+1 vs y là 1 số tự nhiên .
a: \(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)
\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)=\left(\dfrac{x-9}{10}+1\right)+\left(\dfrac{x-10}{11}+1\right)+\left(\dfrac{x-11}{12}+1\right)\)
=>x+1=0
hay x=-1
c: |x-2|=13
=>x-2=13 hoặc x-2=-13
=>x=15 hoặc x=-11
d: \(\Leftrightarrow3\left|x-2\right|+4\left|x-2\right|=2-\dfrac{1}{3}=\dfrac{5}{3}\)
=>7|x-2|=5/3
=>|x-2|=5/21
=>x-2=5/21 hoặc x-2=-5/21
=>x=47/21 hoặc x=37/21
8/9 : ( 2 - 3 x y ) = 5/3
( 2 - 2/3 x y ) : 4 + 7/12 = 11/12
3 : ( 2 x y - 6/15 ) = 1 và 1/2 ( k biết ghi hỗn số nên ghi vậy cho dễ hiểu ạ )
2 - 1/5 x ( y : 7/2 + 1 ) = 1/2
2 và 3/5 x ( 5 : y ) - 3/4 = 0
7/12 : y + 4/9 x 5/8 = 0
4/15 + 2 : ( y + 2/5 ) = 1/5
\(\dfrac{8}{9}\) : ( 2 - 3 \(\times\) y) = \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{9}\) : \(\dfrac{5}{3}\)
2 - 3 \(\times\) y = \(\dfrac{8}{15}\)
3 \(\times\) y = 2 - \(\dfrac{8}{15}\)
3 \(\times\) y = \(\dfrac{22}{15}\)
y = \(\dfrac{22}{15}\) : 3
y = \(\dfrac{22}{45}\)
1,Cho 4x/2x+y =8 và 9x+y/35y=243 ( x,y là số tự nhiên ) Tính x.y
2, Tìm các số hữu tỷ x,y biết : 2x=8y+1 và 9y=3x-9
1,Cho 4x/2x+y =8 và 9x+y/35y=243 ( x,y là số tự nhiên ) Tính x.y
2, Tìm các số hữu tỷ x,y biết : 2x=8y+1 và 9y=3x-9