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\(B=4x-4y+5xy=4.\left(x-y\right)+5xy\)
Thay \(x-y=\frac{5}{12};xy=-\frac{1}{3}\)vào B ta có
\(B=4.\frac{5}{12}+5.\frac{-1}{3}=\frac{20}{12}+\frac{-5}{3}=0\)
Vậy B = 0
B=4x-4y+5xy
B=4.5/12+5.-1/3
B=4.5/12+1+4.-1/3
B=4.(5/12+-1/3)+1
B=4.1/12+1
B=1/3+1
B=4/3
\(A=\left(\frac{1}{10}-1\right)\left(\frac{1}{11}-1\right)\left(\frac{1}{12}-1\right)...\left(\frac{1}{100}-1\right)\)
\(-A=\left(1-\frac{1}{10}\right)\left(1-\frac{1}{11}\right)\left(1-\frac{1}{12}\right)...\left(1-\frac{1}{100}\right)\)
\(-A=\frac{9}{10}\cdot\frac{10}{11}\cdot\frac{11}{12}\cdot...\cdot\frac{99}{100}\)
\(-a=\frac{9}{100}\)
\(A=-\frac{9}{100}\)
Bài 1.
Ta có: \(\frac{a}{b}+\frac{-a}{b+1}=\frac{a}{b}-\frac{a}{b+1}=a\left(\frac{1}{b}-\frac{1}{b+1}\right)=a\left(\frac{b+1-b}{b\left(b+1\right)}\right)=a\left(\frac{1}{b\left(b+1\right)}\right)=\frac{a}{b\left(b+1\right)}\)
=> A là đáp án đúng
Bài 2. Ta có:
B = 4x - 4y + 5xy
B= 4x - 4y + 4xy + xy
B = 4(x - y + xy) + xy
B = 4.(5/12 - 1/3) - 1/3
B = 4.1/12 - 1/3
B = 1/3 - 1/3 = 0
\(\begin{array}{l}B = \left( {\frac{{ - 3}}{{13}}} \right) + \frac{{16}}{{23}} + \left( {\frac{{ - 10}}{{13}}} \right) + \frac{5}{{11}} + \frac{7}{{23}}\\ = \left[ {\left( {\frac{{ - 3}}{{13}}} \right) + \left( {\frac{{ - 10}}{{13}}} \right)} \right] + \left[ {\frac{{16}}{{23}} + \frac{7}{{23}}} \right] + \frac{5}{{11}}\\ = - 1 + 1 + \frac{5}{{11}}\\ = \frac{5}{{11}}\end{array}\)
\(B=4\left(x-y\right)+5xy=4\cdot\dfrac{5}{12}+5\left(-\dfrac{1}{3}\right)=\dfrac{5}{3}-\dfrac{5}{3}=0\)
Thay 6 = x+y
A= \(\frac{4x+y}{3x+\left(x+y\right)}+\frac{4y+x-2.\left(x+y\right)}{3y-\left(x+y\right)}\)
A= \(\frac{4x+y}{4x+y}+\frac{4y+x-2x-2y}{3y-x-y}\)
A=\(1+\frac{2y-x}{2y-x}\)= 1+1=2
B = 4x - 4y + 5xy
B = 4 (x-y ) + 5xy
Thay x-y = \(\frac{5}{12}\) ; xy = \(-\frac{1}{3}\) vào biểu thức B ta có :
B = 4 . \(\frac{5}{12}\) + 5 . \(-\frac{1}{3}\) = \(\frac{20}{12}+-\frac{5}{3}=0\)
Vậy B = 0