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\(\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right)=\frac{a^2-1}{a^2-a}=\frac{a+1}{a}\)
ở phàn a+/a thiếu số 1 nhé
\(\frac{1}{a+1}+\frac{2}{a^2-1}=\frac{a-1+2}{a^2-1}=\frac{1}{a-1}\)
=> K =\(\frac{a^2-1}{a}\)
đkxđ: a khác +-1
b, thay vào mà tình
a/ \(K=\left(\frac{a}{a-1}-\frac{1}{a^2-a}\right):\left(\frac{1}{a+1}+\frac{2}{a^2-1}\right)\)
\(=\left(\frac{a}{a-1}-\frac{1}{a\left(a-1\right)}\right):\left(\frac{1}{a+1}+\frac{2}{\left(a-1\right)\left(a+1\right)}\right)\)
\(=\frac{a^2-1}{a\left(a-1\right)}:\frac{a-1+2}{\left(a-1\right)\left(a+1\right)}\)
\(=\frac{\left(a-1\right)\left(a+1\right)}{a\left(a-1\right)}.\frac{\left(a-1\right)\left(a+1\right)}{a-1}\)
\(=\frac{a+1}{a}.a+1\)
\(=\frac{\left(a+1\right)^2}{a}\)
b, Thay a=1/2
\(\Rightarrow\frac{\left(\frac{1}{2}+1\right)^2}{\frac{1}{2}}=\frac{\frac{9}{4}}{\frac{1}{2}}=\frac{9}{2}\)
(1+1/3)(1+1/8)(1+1/15)...(1+1/9603)=4/3 . 9/8 . 16/15 ... 9604/9603
= (2.2)/(1.3) . (3.3)/(2.4) . (4.4)/(3.5) ... (98.98)/(97.99)
=(2.2.3.3.4.4...98.98)/(1.3.2.4.3.5...97.99)
=(2.3.4...98)/(1.2.3...97) . (2.3.4..98)/(3.4.5...99)
=98/1 .2/99 =169/99 .
a) Đk: x > 0 và x khác +-1
Ta có: A = \(\left(\frac{x+1}{x}-\frac{1}{1-x}-\frac{x^2-2}{x^2-x}\right):\frac{x^2+x}{x^2-2x+1}\)
A = \(\left[\frac{\left(x-1\right)\left(x+1\right)+x-x^2+2}{x\left(x-1\right)}\right]:\frac{x\left(x+1\right)}{\left(x-1\right)^2}\)
A = \(\frac{x^2-1+x-x^2+2}{x\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{x\left(x+1\right)}\)
A = \(\frac{x+1}{x}\cdot\frac{x-1}{x\left(x+1\right)}=\frac{x-1}{x^2}\)
b) Ta có: A = \(\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> 1/x - 1/2 = 0 <=> x = 2 (tm)
Vậy MaxA = 1/4 <=> x = 2
\(A-1=\frac{1}{1.2}+\frac{1}{2.3}..+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}\)\(=\frac{99}{100}\)
\(A=1+\frac{99}{100}=\frac{199}{100}\)
ĐKXĐ \(\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)
với ĐKXĐ ta có
=\(\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x+1\right)\left(x-1\right)}\right):\frac{2x}{7\left(x-1\right)}\)
=\(\frac{4x}{\left(x+1\right)\left(x-1\right)}\times\frac{7\left(x-1\right)}{2x}\)
=\(\frac{14}{x+1}\)
b, x=6(t/m)
khi x=6 thì A=\(\frac{14}{6+1}=2\)
c,A=7<=>\(\frac{14}{x+1}=7\)
\(\Leftrightarrow7x+7=14\)
\(\Leftrightarrow7x=7\Leftrightarrow x=1\left(loại\right)\)
Vậy ko có giá trị x để A=7
\(1+\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=2-\frac{1}{100}\)
\(=\frac{199}{100}\)
Gọi biểu thức là A
A=1+1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A-1=1/2+1/2.3+1/3.4+...+1/98.99+1/99.100
A-1=1-1/2+1/2-1/3+1/3-1/4+...+/198-1/99+1/99-1/100
A-1=1-1/100
A-1=99/100
A=99/100+1
A=199/100
Ta có:
\(A=\frac{1}{358}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{358}\right).2.\frac{1}{297}-7.\frac{1}{358}-\frac{3}{297}.\frac{1}{358}\)
\(=\frac{1}{358}.\left(7+\frac{1}{297}-7-\frac{3}{297}\right)-\left(4-\frac{1}{358}\right).\frac{2}{297}\)
\(=\frac{1}{358}.\left(-\frac{2}{297}\right)-\frac{2}{297}.\left(4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(\frac{1}{358}+4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(-4\right)\)
\(=\frac{8}{297}\)
Vậy giá trị biểu thức A là \(\frac{8}{297}\)
đặt 1/6 chung rồi giải