965 + 546

= 1511

Hok tốt

 965 + 546 

= 1511

hok giỏi nha

A=\(\frac{1}{2}:1\frac{1}{2}:1\frac{1}{3}:1\frac{1}{4}:1\frac{1}{5}:...:1\frac{1}{2000}\)

<=>A=\(\frac{1}{2}:\frac{3}{2}:\frac{4}{3}:\frac{5}{4}:...:\frac{2001}{2000}\)

<=> A=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{2000}{2001}\)

<=> A=\(\frac{1}{2001}\)

 Vậy A=\(\frac{1}{2001}\)

\(A=\frac{1}{2}:1\frac{1}{2}:1\frac{1}{3}:...:1\frac{1}{2000}\)

\(A=\frac{1}{2}:\frac{3}{2}:\frac{4}{3}:...:\frac{2001}{2000}\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2000}{2001}\)

\(A=\frac{1.2.3.....2000}{2.3.4.....2001}\)

\(A=\frac{1}{2001}\)

Vậy : \(A=\frac{1}{2001}\)

\(A=\frac{x^2-x+2}{x^2}\)

\(A=\frac{x^2}{x^2}-\frac{x}{x^2}+\frac{2}{x^2}\)

\(A=1-\frac{1}{x}+2\cdot\left(\frac{1}{x}\right)^2\)

Đặt \(\frac{1}{x}=a\)

\(A=1-a+2a^2\)

\(A=2\left(a^2-\frac{a}{2}+\frac{1}{2}\right)\)

\(A=2\left(a^2-2\cdot a\cdot\frac{1}{4}+\frac{1}{16}+\frac{7}{16}\right)\)

\(A=2\left[\left(a-\frac{1}{4}\right)^2+\frac{7}{16}\right]\)

\(A=2\left(a-\frac{1}{4}\right)^2+\frac{7}{8}\ge\frac{7}{8}\forall a\)

Dấu "=" xảy ra \(\Leftrightarrow a=\frac{1}{4}\Leftrightarrow\frac{1}{x}=\frac{1}{4}\Leftrightarrow x=4\)

Đặt A=1+3+3²+....+3²⁰⁰⁰

=> 3A=3+3²+3³+.....+3²⁰⁰¹

=> 3A-A=2A=3²⁰⁰¹-1

=> A=(3²⁰⁰¹-1)/2

=> S=(3²⁰⁰¹-1)/2(1-3²⁰⁰¹)

=> S=-1/2

Đúng thì tk cho mình nha. 

Đặt \(A=1+3+3^2+3^3+...+3^{2000}\)

\(\Rightarrow3A=3+3^2+3^3+...+3^{2001}\)

\(\Rightarrow3A-A=3^{2001}-1\)

\(\Rightarrow2A=3^{2001}-1\)

\(\Rightarrow A=\frac{3^{2001}-1}{2}\)

Vậy \(S=\frac{\frac{3^{2001}-1}{2}}{1-3^{2001}}\)\(=\frac{3^{2001}-1}{2}\cdot\frac{1}{1-3^{2001}}=\frac{3^{2001}-1}{2\cdot\left(1-3^{2001}\right)}=-\frac{1}{2}\)

\(192-\left(x^2-1\right)\left(x^2+4x+3\right)=0\)

\(\Leftrightarrow192-\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)=0\)

\(\Leftrightarrow192-\left[\left(x-1\right)\left(x+3\right)\right]\left[\left(x+1\right)\left(x+1\right)\right]=0\)

\(\Leftrightarrow192-\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)

Đặt \(x^2+2x-3=a\)

\(pt\Leftrightarrow192-a\left(a+4\right)=0\)

\(\Leftrightarrow192-a^2-4a=0\)

\(\Leftrightarrow-a^2-16a+12a+192=0\)

\(\Leftrightarrow-a\left(a+16\right)+12\left(a+16\right)=0\)

\(\Leftrightarrow\left(a+16\right)\left(-a+12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-16\\a=12\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2x-3=-16\\x^2+2x-3=12\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2x+13=0\\x^2+2x-15=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+2x+1+12=0\\x^2+5x-3x-15=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=-12\\x\left(x+5\right)-3\left(x+5\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\in\varnothing\\\left(x+5\right)\left(x-3\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)

Vậy.....

giải phương trình: 192-(x²-1)(x²+4x+3)=0

\(Ta\)\(có\)

\(P\left(x\right)=-F\left(x\right)\)

\(\Leftrightarrow-x^3+2x^2+x-1=\left(x^3-2x^2+3x+5\right)\)

\(\Leftrightarrow-x^3+2x^2+x-1=-x^3+2x^2-3x-5\)

\(\Leftrightarrow x-1=-3x-5\)

\(\Leftrightarrow4x=-4\)

\(\Leftrightarrow x=-1\)

\(Vậy\)\(......\)

sorry \(-x^3+2x^2+x-1=-\left(x^3-2x+3x+5\right)\)

mik thiếu dấu trừ nha