a)\(B=\dfrac{a^2}{a-1}+\left(\dfrac{a}{a^2-1}+\dfrac{1}{a-a^3}\right):\dfrac{1-a}{a+a^3}\)

\(B=\dfrac{a^2}{a-1}+\dfrac{a^2-1}{a\left(a^2-1\right)}.\dfrac{a+a^3}{1-a}\)

\(B=\dfrac{a^2}{a-1}+\dfrac{1}{a}.\dfrac{a\left(1+a^2\right)}{1-a}\)

\(B=\dfrac{a^2}{a-1}+\dfrac{1+a^2}{1-a}\)

\(B=\dfrac{a^2}{a-1}+\dfrac{-1-a^2}{a-1}\)

\(B=\dfrac{-1}{a-1}\)

b)\(2a^2=a\Leftrightarrow2a^2-a=0\Leftrightarrow a\left(2a-1\right)=0\)<=>a=0 hoặc 2a-1=0

<=>a=0 hoặc a=1/2

TH1: a=0 => \(B=\dfrac{-1}{a-1}=\dfrac{-1}{0-1}=\dfrac{-1}{-1}=1\)

TH2: a=1/2 => \(B=\dfrac{-1}{a-1}=\dfrac{-1}{\dfrac{1}{2}-1}=\dfrac{-1}{-\dfrac{1}{2}}=2\)

c)\(B=\dfrac{-1}{a-1}< 1\Leftrightarrow-1< a-1\Leftrightarrow a>0\)

Vậy B<1 khi a>0

\(4x^2-12x+5=0\Leftrightarrow4x^2-10x-2x+5=0\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy ...

ok ,đồng ý luôn nhưng thi lớp mấy zợ

\(P\left(x\right)=Q\left(x\right)\Leftrightarrow4x^3-3x^2+4x-4=4x^3-x^2-3x-4\)

<=>\(4x^3-3x^2+4x-4-4x^3+x^2+3x+4=0\)

<=>\(-2x^2+7x=0\Leftrightarrow x\left(-2x+7\right)=0\)<=>x=0 hoặc -2x+7=0

<=>x=0 hoặc x=3,5

Vậy P(x)=Q(x) khi x=0 hoặc x=3,5

Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};\dfrac{1}{4^2}< \dfrac{1}{3.4};...;\dfrac{1}{n^2}< \dfrac{1}{\left(n-1\right)n}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)

mà \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right)n}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\)

\(=1-\dfrac{1}{n}< 1\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}< 1\) (đpcm)

=>\(2a=8\Rightarrow a=4\)

\(a-b+c=11;a-b-c=-1\Rightarrow\left(a-b+c\right)-\left(a-b-c\right)=11-\left(-1\right)\)

\(\Rightarrow a-b+c-a+b+c=11+1\Rightarrow2c=12\Rightarrow c=6\)

\(a+b-c=4+b-6=b-2=-3\Rightarrow b=-1\)

Vậy a=4;b=-1;c=6

1.Yesterday was Hoa’s thirteenth birthday.

2.Because she ate a lot of food, fruits and cakes.

3.Yes, her aunt was.

4.When I have a stomathache I go to the doctor.

\(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)

<=>\(\dfrac{x-1}{2009}-1+\dfrac{x-2}{2008}-1=\dfrac{x-3}{2007}-1+\dfrac{x-4}{2006}-1\)

<=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)

<=>\(\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)

<=>\(\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)

Vì \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên x-2010=0 <=>x=2010

Số bạn chỉ giỏi văn hoặc giỏi toán và số bạn giỏi cả văn lẫn toán là: 33+31=64(bạn)

Số bạn giỏi cả văn lẫn toán là: 64-53=11 (bạn)

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