Tính giá trị biểu thức
\(B=\left(200^{-2}-1\right)\left(199^{-2}-1\right)\left(198^{-2}-1\right)...\left(101^{-2}-1\right)\)
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\(A=202\left(200^{-2}-1\right)\left(199^{-2}-1\right)\left(198^{-2}-1\right)...\left(101^{-2}-1\right)\)
\(=202\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)\left(\frac{1}{198^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)
\(=-202\left(1-\frac{1}{200^2}\right)\left(1-\frac{1}{199^2}\right)\left(1-\frac{1}{198^2}\right)...\left(1-\frac{1}{101^2}\right)\)
\(=-202\left(\frac{199.201}{200^2}\right).\left(\frac{198.200}{199^2}\right).\left(\frac{197.199}{198^2}\right)...\left(\frac{102.100}{101^2}\right)\)
\(=-202.\frac{199.201.198.200.197.199...100.102}{200^2.199^2.198^2...101^2}\)
\(=-202.\frac{\left(199.198.197...100\right)\left(201.200.199...102\right)}{\left(200.199.198...101\right)\left(200.199.198...101\right)}\)
\(=-202.\frac{1.201}{2.101}=-202.\frac{201}{202}=-201\)
\(C=\left(\dfrac{1}{200^2}-1\right)\left(\dfrac{1}{199^2-1}\right)...\left(\dfrac{1}{101^2-1}\right)\)
\(C=\dfrac{1-200^2}{200^2}.\dfrac{1-199^2}{199^2}.\dfrac{1-198^2}{198^2}...\dfrac{1-101^2}{101^2}\)
\(C=\dfrac{\left(1-200\right)\left(1+200\right)}{200^2}.\dfrac{\left(1-199\right)\left(1+199\right)}{199^2}...\dfrac{\left(1-100\right)\left(1+100\right)}{100^2}.\dfrac{\left(1-101\right)\left(1+101\right)}{101^2}\) \(C=\dfrac{-199.201}{200.200}.\dfrac{-198.200}{199.199}.\dfrac{-197.199}{198.198}...\dfrac{-99.101}{100.100}.\dfrac{-100.102}{101.101}\)
\(C=\dfrac{199.201}{200.200}.\dfrac{198.200}{199.199}.\dfrac{197.199}{198.198}...\dfrac{99.101}{100.100}.\dfrac{100.102}{101.101}\)
\(\Rightarrow C=\dfrac{200}{2.101}=\dfrac{201}{202}\)
Câu 2 mik chịu r sorry:(
S= 1/199 + 2/198 + ... + 198/2 + 199/1
S= (1/199 + 1) + (2/198 + 1)+ ... + (198/2 + 1) +1
S= 200/200 + 200/199 + 200/198 + ... + 200/2
S= 200.(1/200 + 1/199 + ... + 1/2)
Suy ra , B=(1/2 + 1/3 + ... +1/200) : 200.(1/2 + 1/3 + ... + 1/200)
B=1 : 200 = 1/200
a) \(\left(\dfrac{3}{4}\right)^{-2}\cdot3^2\cdot12^0=16\)
b) \(\left(\dfrac{1}{12}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-2}=27\)
c) \(\left(2^{-2}\cdot5^2\right)^{-2}:\left(5\cdot5^{-5}\right)=16\)
a) \(\left(-5\right)^{-1}=-\dfrac{1}{5}\)
b) \(2^0\cdot\left(\dfrac{1}{2}\right)^{-5}=1\cdot32=32\)
c) \(6^{-2}\cdot\left(\dfrac{1}{3}\right)^{-3}:2^{-2}\)
\(=\dfrac{1}{36}\cdot27:\dfrac{1}{4}\)
\(=\dfrac{27\cdot4}{36}=3\)
1: A=4x^2+12x+9-4x^2+4x-1-6x=10x+8
Khi x=201 thì A=10*201+8=2018
2: B=4x^2+20x+25-4x^2+12=20x+37
Khi x=1/20 thì B=1+37=38
1, \(A=\left(2x+3\right)^2-\left(2x-1\right)^2-6x\)
\(A=\left[\left(2x+3\right)+\left(2x-1\right)\right]\left[\left(2x+3\right)-\left(2x-1\right)\right]-6x\)
\(A=\left(2x+3+2x-1\right)\left(2x+3-2x+1\right)-6x\)
\(A=4\left(4x+2\right)-6x\)
\(A=16x+8-6x\)
\(A=10x+8\)
Thay \(x=201\) vào A ta có:
\(A=10\cdot201+8=2010+8=2018\)
Vậy: ....
2, \(B=\left(2x+5\right)^2-4\left(x+3\right)\left(x-3\right)\)
\(B=\left(2x+5\right)^2-4\left(x^2-9\right)\)
\(B=4x^2+20x+25-4x^2+36\)
\(B=20x+61\)
Thay \(x=\dfrac{1}{20}\) vào B ta có:
\(B=20\cdot\dfrac{1}{20}+61=1+61=62\)
Vậy: ...
\(M=\left(7-2x\right)\left(4x^2+14x+49\right)-\left(64-8x^3\right)\)
\(M=\left(7-2x\right)\left[\left(2x\right)^2+2x\cdot7+7^2\right]-\left(64-8x^3\right)\)
\(M=\left[7^3-\left(2x\right)^3\right]-\left(64-8x^3\right)\)
\(M=343-8x^3-64+8x^3\)
\(M=279\)
Vậy M có giá trị 279 với mọi x
\(P=\left(2x-1\right)\left(4x^2-2x+1\right)-\left(1-2x\right)\left(1+2x+4x^2\right)\)
\(P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3\)
\(P=16x^3-8x^2+4x-2\)
Thay \(x=10\) vào P ta có:
\(P=16\cdot10^3-8\cdot10^2+4\cdot10-2=15238\)
Vậy P có giá trị 15238 tại x=10
a: M=343-8x^3-64+8x^3=279
b: P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3
=16x^3-8x^2+4x-2
=16*10^3-8*10^2+4*10-2=15238
Áp dụng tính chất a2 - b2 = a2 - ab + ab - b2 = a(a - b) + b(a - b) = (a + b)(a - b)
B =\(\left(200^{-2}-1\right)\left(199^{-2}-1\right)...\left(101^{-2}-1\right)=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)
\(=\frac{1-200^2}{200^2}.\frac{1-199^2}{199^2}...\frac{1-101^2}{101^2}=\frac{1^2-200^2}{200^2}.\frac{1^2-199^2}{199^2}....\frac{1^2-101^2}{101^2}\)
\(=\frac{\left(1-200\right)\left(1+200\right)}{200^2}.\frac{\left(1-199\right)\left(1+199\right)}{199^2}...\frac{\left(1-101\right)\left(1+101\right)}{101^2}\)
\(=-\left(\frac{199.201}{200^2}.\frac{198.200}{199^2}...\frac{100.102}{101^2}\right)=-\frac{199.201.198.200..100.102}{200.200.199.199...101.101}\)
\(=-\frac{\left(199.198...100\right)\left(201.200...102\right)}{\left(200.199...101\right).\left(200.199...101\right)}=-\frac{100.201}{200.101}=-\frac{201}{202}\)
Bài giải
\(B=\left(200^{-2}-1\right)\left(199^{-2}-1\right)\left(198^{-2}-1\right)...\left(101^{-2}-1\right)\)
\(B=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)\left(\frac{1}{198^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)
\(B=\left[\left(\frac{1}{200}\right)^2-1^2\right]\left[\left(\frac{1}{199}\right)^2-1^2\right]\left[\left(\frac{1}{198}\right)^2-1^2\right]...\left[\left(\frac{1}{101}\right)^2-1^2\right]\)
\(B=\left(\frac{1}{200}+1\right)\left(\frac{1}{200}-1\right)\left(\frac{1}{199}+1\right) \left(\frac{1}{199}-1\right)..\left(\frac{1}{101}-1\right)\left(\frac{1}{101}+1\right)\)
\(B=\frac{201}{200}\cdot\frac{-199}{200}\cdot\frac{200}{199}\cdot\frac{-198}{199}\cdot...\cdot\frac{-100}{101}\cdot\frac{102}{101}\)
\(B=\frac{201\cdot\left(-199\right)\cdot200\cdot\left(-198\right)\cdot...\cdot\left(-100\right)\cdot102}{200\cdot200\cdot199\cdot199\cdot...\cdot101\cdot101}=\frac{100\cdot201}{200\cdot101}=\frac{201}{202}\)