a) \(x^8+x+1\)

\(=x^8+x^7-x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+1\)

\(=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

c) \(x^4+4y^4=x^4+4x^2y^2+4y^4-4x^2y^2=\left(x^4+4x^2y^2+4y^4\right)-4x^2y^2\)

\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)

d) \(x^4+4=x^4+4x^2+4-4x^2=\left[\left(x^2\right)^2+2\cdot x^2\cdot2+2^2\right]-4x^2\)

\(=\left(x^2+2\right)^2-\left(2x\right)^2=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)

\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)

*) \(4x^2-1=0\)

\(\Rightarrow4x^2=1\Rightarrow x^2=\dfrac{1}{4}\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

*) \(2x^2+0,82=1\)

\(\Rightarrow2x^2=1-0,82=\dfrac{9}{50}\)

\(\Rightarrow x^2=\dfrac{9}{100}\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{3}{10}\end{matrix}\right.\)

*) \(\left(3x-\dfrac{1}{4}\right)\left(x+\dfrac{1}{2}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x-\dfrac{1}{4}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{12}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\left(x+1\right)\left(x+4\right)=5\sqrt{x^2+5x+28}\)

\(Pt\Leftrightarrow x^2+5x+4-5\sqrt{x^2+5x+28}=0\)

Đặt \(t=\sqrt{x^2+5x+28}\) \(\left(t>0\right)\)

Ta có: \(t^2=x^2+5x+28\)

\(\Leftrightarrow x^2+5x+4=t^2-24\)

Thay vào pt ta được:

\(t^2-24-5t=0\)

\(\Leftrightarrow t^2-8t+3t-24=0\)

\(\Leftrightarrow\left(t-8\right)\left(t+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=8\left(tm\right)\\t=-3\left(loai\right)\end{matrix}\right.\)

*) \(t=8\Leftrightarrow x^2+5x-36=0\)

\(\Leftrightarrow x^2+9x-4x-36=0\)

\(\Leftrightarrow x\left(x+9\right)-4\left(x+9\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=4\end{matrix}\right.\)(tm)

Vậy.................

\(=\frac{7}{2}x\frac{1269}{46}x\frac{1}{17}x\frac{22}{9}=\frac{7x1269x22}{2x46x17x9}=\frac{10857}{782}\)

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

\(=\left|x+1\right|+\left|1-x\right|\)

Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(\left|x+1\right|+\left|1-x\right|\ge\left|x+1+1-x\right|=2\)

Dấu ''='' xảy ra khi \(x\le1\)

Con bitch hack nick Linh-sama là con này chứ ai: Đinh Hải Ngọc

Đã hack nik lại còn sử dụng 1 cách ngu vồn

chậc, để t tiễn m chặng đường cuối về địa phủ nhé :)

A = (2n - 1 + 1)x n : 2 =2n x n : 2 =n^2

y+3,5+y=24,5=>2y+3,5=24,5=>2y=24,5-3,5=>2y=21=>y=21:2=>y=10,5

Vay : y=10,5

c, 

x(y-2)(x^2-9)=0 

suy ra : x=0

TH1 : y-2=0=>y=0+2=>y=2

TH2:z^2-9=0=>z^2=0+9=>z^2=9=>z=+-3

Vay x=0 ; y=2 va z=3

\(x^2+4x+2+\left|x\right|+\sqrt{4-x^2}=0\)

\(\Leftrightarrow-x^2-4x-2=\left|x\right|+\sqrt{4-x^2}\)

\(\Leftrightarrow VP^2=\left(\left|x\right|+\sqrt{4-x^2}\right)^2=4+2\left|x\right|\sqrt{4-x^2}\)

\(\Leftrightarrow VP^2\ge4\Rightarrow VP\ge2\)

Do vậy: \(VT\ge2\)

\(\Leftrightarrow-x^2-4x-2\ge2\)

\(\Leftrightarrow-x^2-4x-4\ge0\)

\(\Leftrightarrow\left(x+2\right)^2\le0\) \(\Leftrightarrow x=-2\)

Vậy x = -2 là nghiệm của pt

gọi 3 góc A,B,C lần lượt là a,b,c (a,b,c > 0)

Theo đề ta có: \(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{7}\) và a+b+c = 180

Áp dụng tccdts=nhau có:

\(\dfrac{a}{3}=\dfrac{b}{5}=\dfrac{c}{7}=\dfrac{a+b+c}{3+5+7}=\dfrac{180}{15}=12\)

=> a = 12 . 3 = 36

Vậy \(\widehat{A}=36^o\)