a) - Thu gọn đa thức P(x):

P(x)=2+3x2−3x3+5x4−2x−x3+7x5=2+3x2−(3x3+x3)+5x4−2x+7x5P(x)=2+3x2−3x3+5x4−2x−x3+7x5=2+3x2−(3x3+x3)+5x4−2x+7x5 =2+3x2−4x3+5x4−2x+7x5

2) 

a) ĐK: \(2x^2-8x-12\ge0\)(1)

Nhân 2 cả hai vế ta có:

\(2x^2-8x-12=2\sqrt{2x^2-8x-12}\)

Đặt: \(\sqrt{2x^2-8x-12}=t\left(t\ge0\right)\)

Ta có phương trình: \(t^2=2t\Leftrightarrow\orbr{\begin{cases}t=0\\t=2\end{cases}}\)(tm)

+) Với t=0  ta có:\(\sqrt{2x^2-8x-12}=0\Leftrightarrow2x^2-8x-12=0\Leftrightarrow x^2-4x-6=0\Leftrightarrow\orbr{\begin{cases}x=2+\sqrt{10}\\x=2-\sqrt{10}\end{cases}}\)( thỏa mãn đk (1))

+) Với t=2 ta có: \(\sqrt{2x^2-8x-12}=2\Leftrightarrow2x^2-8x-12=4\Leftrightarrow x^2-4x-8=\Leftrightarrow\orbr{\begin{cases}x=2+2\sqrt{3}\\x=2-2\sqrt{3}\end{cases}}\)( THỎA MÃN đk (1))

vậy ...

b) pt <=> \(\left(4x+1\right)\left(3x+2\right)\left(12x-1\right)\left(x+1\right)=4\)

<=> \(\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)=4\)

Đặt :\(12x^2+11x+2=t\)

Ta có pt: \(t\left(t-3\right)=4\Leftrightarrow t^2-3t-4=0\Leftrightarrow\orbr{\begin{cases}t=4\\t=-1\end{cases}}\)

Với t=4 ta có: ....

Với t=-1 ta có:...

Em tự làm tiếp nhé

458 + 524

= 980

hok tốt

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