Thực hiện các phép tính sau:
a) m 2 − 5 m + 6 m 2 + 7 m + 12 : m 2 − 6 m + 9 m 2 + 4 m với m ≠ − 4 ; ± 3 ; 0 ;
b) u 2 + 4 uv + 4 v 2 2 u 2 − 2 uv + 2 v 2 : 4 u − 8 v 6 u 3 − 6 v 3 với u ≠ − v và u ≠ 2 v
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) Ta có:
\(A=-3\cdot7\cdot\left(-2\right)\cdot\left(-13\right)\)
\(A=-21\cdot26\)
\(A=-546\)
\(B=-1\cdot\left(-2\right)\cdot\left(-3\right)\cdot\left(-4\right)\cdot5\)
\(B=2\cdot12\cdot5\)
\(B=2\cdot60\)
\(B=120\)
Mà: \(120>-546\)
\(\Rightarrow B>A\)
a) M = -5/7 . 2/11 + -5/7 . 9/11 + 12/7
M = -5/7 . ( 2/11 + 9/11 ) + 12/7
M = -5/7 . 1 + 12/7
M = -5/7 + 12/7
M = 1
b) N = 6/7 + 5/8 : 5 - 3/16 . ( -2 ) mũ 2
N = 6/7 + 1/8 - 3/16 . 4
N = 48/56 + 7/56 - 3/4
N = 55/56 - 42/56
N = 13/56
a. M=-1^2+2^2-3^2+4^2-...-99^2+100^2.
M=(2-1)(2+1)+(4-3)(4+3)+...+(100-99)(100+99)
M=3+7+...+199
=>2M=3+7+...+199+3+7+...+199 (198 số)
=(3+199)+(7+195)+...+(199+3) (99 cặp)
=202.99
=19998
=>M=19998:2=9999
\(M=\left(\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right)\left(\sqrt{6}+1\right)\)
\(=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+1\right)\)
\(=\left(\sqrt{6}-11\right)\left(\sqrt{6}+1\right)\)
\(=6+\sqrt{6}-11\sqrt{6}-11=-5-10\sqrt{6}\)
\(M=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+1\right)\)
\(M=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}+2\right)\left(\sqrt{6}-2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+1\right)\)
\(M=\left[\dfrac{15\left(\sqrt{6}-1\right)}{6-1}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+1\right)\)
\(M=\left[3\left(\sqrt{6}-1\right)+2\left(\sqrt{6}+2\right)-4\left(3+\sqrt{6}\right)\right]\left(\sqrt{6}+1\right)\)
\(M=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\cdot\left(\sqrt{6}+1\right)\)
\(M=\left(5\sqrt{6}-4\sqrt{6}+1-12\right)\left(\sqrt{6}+1\right)\)
\(M=\left(\sqrt{6}-11\right)\left(\sqrt{6}+1\right)\)
\(M=6+\sqrt{6}-11\sqrt{6}-11\)
\(M=-10\sqrt{6}-5\)
M=3/1.2-5/2.3+7/3.4-9/4.5+11/5.6-13/6.7+15/7.8+17/8.9
=(1/1.1+2/1.2)-(2/2.3+3/2.3)+(3/3.4+4/3.4)-(4/4.5+5/4.5)+...+(8/8.9+9/8.9)(phần ... là làm tương tự nhé)
=1/2+1-(1/3+1/2)+(1/4+1/3)-(1/5+1/4)+...+(1/9+1/8)(phần ... là làm tương tự nhé)
=1+(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+...+(1/8-1/8)-1/9
=1+0+0+0+...+0-1/9
=1-1/9
=8/9
Bài 2:
1: \(A=\left(x+2\right)\left(x^2-2x+4\right)+2\left(x+1\right)\left(1-x\right)\)
\(=\left(x+2\right)\left(x^2-x\cdot2+2^2\right)-2\left(x+1\right)\left(x-1\right)\)
\(=x^3+2^3-2\left(x^2-1\right)\)
\(=x^3+8-2x^2+2=x^3-2x^2+10\)
\(B=\left(2x-y\right)^2-2\left(4x^2-y^2\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y\right)^2-2\cdot\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)^2+4\left(y+2\right)\)
\(=\left(2x-y-2x-y\right)^2+4\left(y+2\right)\)
\(=\left(-2y\right)^2+4\left(y+2\right)\)
\(=4y^2+4y+8\)
2: Khi x=2 thì \(A=2^3-2\cdot2^2+10=8-8+10=10\)
3: \(B=4y^2+4y+8\)
\(=4y^2+4y+1+7\)
\(=\left(2y+1\right)^2+7>=7>0\forall y\)
=>B luôn dương với mọi y
Bài 1:
5: \(x^2\left(x-y+1\right)+\left(x^2-1\right)\left(x+y\right)\)
\(=x^3-x^2y+x^2+x^3+x^2y-x-y\)
\(=2x^3-x+x^2-y\)
6: \(\left(3x-5\right)\left(2x+11\right)-6\left(x+7\right)^2\)
\(=6x^2+33x-10x-55-6\left(x^2+14x+49\right)\)
\(=6x^2+23x-55-6x^2-84x-294\)
=-61x-349
Bài 1:
Ta có: \(\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{6}-1\right)\left(\dfrac{1}{10}-1\right)\cdot...\cdot\left(\dfrac{1}{45}-1\right)\)
\(=\dfrac{-2}{3}\cdot\dfrac{-5}{6}\cdot\dfrac{-9}{10}\cdot...\cdot\dfrac{-44}{45}\)
\(=\dfrac{-2}{3}\cdot\dfrac{-5}{6}\cdot\dfrac{-9}{10}\cdot\dfrac{-14}{15}\cdot\dfrac{-20}{21}\cdot\dfrac{-27}{28}\cdot\dfrac{-35}{36}\cdot\dfrac{-44}{45}\)
\(=\dfrac{11}{27}\)
Câu 2:
B=1+1/2+1/3+....+1/2010
=(1+1/2010)+(1/2+1/2009)+(1/3+1/2008)+...(1/1005+1/1006)
= 2011/2010+2011/2.2009+2011/3.2008+...+2011/1005.1006
=2011.(1/2010+.....1/1005.1006)
Vậy B có tử số chia hết cho 2011 (đpcm).
Câu 3:
\(P=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}....\dfrac{98}{99}\\ P< \dfrac{3}{4}.\dfrac{5}{6}.\dfrac{6}{7}....\dfrac{99}{100}\\ P^2< \dfrac{2}{100}\)
Mà
\(\dfrac{2}{100}=\dfrac{1}{50}< \dfrac{1}{49}\\ \Rightarrow P< \dfrac{1}{7}\)
a,5/9+(-3/5) +( -5/9)
=5/9+(-5/9)+(-3/5)
=0+(-3/5)
=-3/5
b, b,-3/7. 2/11- (-3/7).9/11
=-3/7.(2/11-9/11)
=-3/7.(-7/11)
=3/11
c,,2/3 + 5/6 : 5 - 1/18 . (-3)
=2/3+(1/6-(-1/6))
=2/3+1/3
=3/3
=1
tick nếu đúng nha
a) m 2 − 2 m m 2 − 9 b) 3 4 ( u − 2 v ) ( u + v )