a) 

<=> \(3x-12x^2+12x^2-6x=9\)

<=> \(-3x=9\)

<=> \(x=-3\)

b)

<=> \(6x-24x^2-12x+24x^2=6\)

<=> \(-6x=6\)

<=> \(x=-1\)

c) 

<=> \(6x-4-3x+6=1\)

<=> \(3x+2=1\)

<=> \(x=-\frac{1}{3}\)

d) 

<=> \(9-6x^2+6x^2-3x=9\)

<=> \(-3x=0\)

<=> \(x=0\)

e) KO HIỂU ĐỀ

f) 

<=> \(4x^2-8x+3-\left(4x^2+9x+2\right)=8\)

<=> \(-17x+1=8\)

<=> \(x=-\frac{7}{17}\)

g) 

<=> \(-6x^2+x+1+6x^2-3x=9\)

<=> \(-2x=8\)

<=> \(x=-4\)

h)

<=> \(x^2-x+2x^2+5x-3=4\)

<=> \(3x^2+4x=7\)

<=> \(\orbr{\begin{cases}x=1\\x=-\frac{7}{3}\end{cases}}\)

a. \(3x\left(1-4x\right)+6x\left(2x-1\right)=9\)

\(\Rightarrow3x-12x^2+12x^2-6x=9\)

\(\Rightarrow-3x=9\)

\(\Rightarrow x=-3\)

b. \(3x\left(2-8x\right)-12x\left(1-2x\right)=6\)

\(\Rightarrow6x-24x^2-12x+24x^2=6\)

\(\Rightarrow-6x=6\)

\(\Rightarrow x=-1\)

c. \(2\left(3x-2\right)-3\left(x-2\right)=1\)

\(\Rightarrow6x-4-3x+6=1\)

\(\Rightarrow3x+2=1\)

\(\Rightarrow3x=-1\)

\(\Rightarrow x=-\frac{1}{3}\)

\(=\left(a^2-1\right)^3-\left(a^6-1\right)\)

\(=a^6-3a^4+3a^2-1-a^6+1\)

\(=-3a^4+3a^2\)

\(=-3a^2\left(a^2-1\right)\)

\(\left(a^2-1\right)^3-\left(a^4+a^2+1\right)\left(a^2-1\right)\)

\(=a^6-3a^4+3a^2-1-\left(a^6-a^4+a^4-a^2+a^2-1\right)\)

\(=a^6-3a^4+3a^2-1-a^6+1\)

\(=-3a^4+3a^2\)

\(=-3a^2\left(a^2-1\right)\)

\(=-3a^2\left(a+1\right)\left(a-1\right)\)

Chú ý 2 điều: \(\cos45^o=\sin45^o=\frac{\sqrt{2}}{2}\) và \(\cos^2a+\sin^2a=1\)

Do đó: 

a) \(A=\cos^252^o.\frac{\sqrt{2}}{2}+\sin^252^o.\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}\left(\cos^252^o+\sin^252^o\right)=\frac{\sqrt{2}}{2}.1=\frac{\sqrt{2}}{2}\)

b) \(B=\frac{\sqrt{2}}{2}.\cos^247^o+\frac{\sqrt{2}}{2}.\sin^247^o=\frac{\sqrt{2}}{2}\left(\cos^247^o+\sin^247^o\right)=\frac{\sqrt{2}}{2}.1=\frac{\sqrt{2}}{2}\)

\(F=2x^2+2xy+5y^2-8x-22y\)

<=> \(2F=4x^2+4xy+10y^2-16x-44\)

\(=\left(4x^2+4xy+y^2-16x+16-8y\right)+9y^2-36y-16\)

\(=\left(2x+y-4\right)^2+\left(3y-6\right)^2-52\ge-52\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x+y-4=0\\3y-6=0\end{cases}}\Leftrightarrow y=2;x=1\)

=> min 2F = -52

=> min F = -26. 

\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-4\left(a^3+b^3+c^3\right)-12abc\)

\(=-3\left(a^3+b^3+c^3\right)+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-12abc\)

\(=-3\left(\left(a^3+b^3+c^3\right)-\left(a+b\right)\left(b+c\right)\left(c+a\right)+4abc\right)\)

XONG NHAAAAA :3333333

\(I=1^2+2^2+3^2+4^2+...+2017^2+2018^2\)

\(I=\left(2^2-1^2\right)+\left(4^2-3^2\right)+...+\left(2018^2-2017^2\right)\)

\(I=\left(1+2\right)\left(2-1\right)+\left(3+4\right)\left(4-3\right)+...+\left(2017+2018\right)\left(2018-2017\right)\)

\(I=1+2+3+4+...+2017+2018\)

\(I=\frac{\left(2018+1\right).2018}{2}=2037171\)

I = 1² + 2² + 3² + ... + 2017²

= 1.1 + 2.2 + 3.3 + ... + 2017.2017

= 1.(2 - 1) + 2.(3 - 1) + 3.(4 - 1) + .... + 2017.(2018 - 1)

= 1.2 + 2.3 + 3.4 + ... + 2017.2018 - (1 + 2 + 3 + 4 + ... + 2017)

= 1.2 + 2.3 + 3.4 + ... + 2017.2018 - 2035153

Đặt K = 1.2 + 2.3 + 3.4 + ... + 2017.2018

=> 3K = 1.2.3 + 2.3.3 + 3.4.3 + .... + 2017.2018.3

=> 3K = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 2017.2018.(2019 - 2016)

=> 3K = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 2017.2018.2019 - 2016.2017.2018

=> 3K = 2017.2018.2019

=> K = 2017.2018.2019 : 3 

=> K = 2739315938

Lại có I = K - 2035153

= 2739315938 - 2035153

= 2 737 280 785

pt => \(2x+5=1-x\)

<=> \(3x=4\)

<=> \(x=\frac{4}{3}\)

VẬY    \(x=\frac{4}{3}\)là no duy nhất

Hermit :) làm nhầm rồi em.

\(\sqrt{2x+5}=\sqrt{1-x}\Leftrightarrow\hept{\begin{cases}1-x\ge0\\2x+5=1-x\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le1\\3x=-4\end{cases}}\Leftrightarrow x=-\frac{4}{3}\)

Vậy x = -4/3

lớp 9 thì mình dùng cách lớp 9 

\(\sqrt{x+2\sqrt{x}-1}=2\left(đk:x\ge1\right)\)

\(< =>x+2\sqrt{x}-1=4\)(bình phương 2 vế)

Đặt \(\sqrt{x}=t\left(t\ge0\right)\)(*)

\(< =>t^2+2t-5=0\)

\(\Delta=2^2-4.\left(-5\right)=4+20=24\)

\(\orbr{\begin{cases}t_1=\frac{-2+2\sqrt{6}}{2}=-1+\sqrt{6}\left(tm\right)\\t_2=\frac{-2-2\sqrt{6}}{2}=-1-\sqrt{6}\left(ktm\right)\end{cases}}\)

Khi đó thế vào * ta được :

 \(\sqrt{x}=\sqrt{6}-1< =>x=7-2\sqrt{6}\left(tmđk\right)\)

Vậy nghiệm của phương trình trên là \(7-2\sqrt{6}\)

ĐK: \(x\ge1\)

\(\sqrt{x+2\sqrt{x-1}}=2\)

<=> \(\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=2\)

<=> \(\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)

<=> \(\sqrt{x-1}+1=2\)

<=> \(\sqrt{x-1}=1\)

<=> x - 1 = 1 

<=> x = 2 thỏa mãn