#)Giải :

\(A=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)

\(A=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{2}{3}+\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\left(-\frac{4}{5}+\frac{4}{5}\right)+\left(\frac{5}{6}-\frac{5}{6}\right)-\frac{6}{7}\)

\(A=0+0+0+0+0-\frac{6}{7}\)

\(A=-\frac{6}{7}\)

A=  \(\frac{221}{210}\)

k tui nha, hậu tạ

sao mày o làm nhanh ả

\(a,\text{ }2^n\text{ x }4=512\)

\(2^n=512\text{ : }4\)

\(2^n=128\)

\(2^n=2^7\)

\(\Rightarrow\text{ }n=7\)

\(b,\left(2n+1\right)^3=125\)

\(\left(2n+1\right)^3=5^3\)

\(\Rightarrow\text{ }2n+1=5\)

\(2n=5-1\)

\(2n=4\)

\(n=4\text{ : }2\)

\(n=2\)

\(a,\)\(đkxđ\Leftrightarrow x\ge0\)và \(x-9\ne0\Rightarrow x\ne9\)

\(A=\frac{6\sqrt{x}}{x-9}-\frac{5\sqrt{x}}{3-\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+3}\)

\(\)\(=\frac{6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{5\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{6\sqrt{x}+5x+15\sqrt{x}+x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{18\sqrt{x}+6x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{6\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{6\sqrt{x}}{\sqrt{x}-3}\)

\(b,\)Để \(A>2\)\(\Rightarrow\frac{6\sqrt{x}}{\sqrt{x}-3}>2\)

\(\Rightarrow\frac{6\sqrt{x}}{\sqrt{x}-3}>\frac{12\sqrt{x}}{x-3}\)

\(\Rightarrow\frac{6\sqrt{x}-12\sqrt{x}}{\sqrt{x}-3}>0\)

\(\Rightarrow\frac{6\sqrt{x}}{\sqrt{x}-3}< 0\)

Vì \(\sqrt{x}\ge0;\)\(6>0\)\(\Rightarrow6\sqrt{x}\ge0\)

\(\Rightarrow\frac{6\sqrt{x}}{\sqrt{x}-3}>0\Leftrightarrow\sqrt{x}-3< 0\)

\(\Rightarrow\sqrt{x}< 3\Rightarrow\sqrt{x}< \sqrt{9}\)\(\Leftrightarrow x< 9\)

Mà \(x\ge0\left(đkxđ\right)\)\(\Rightarrow0\le x< 9\)

\(2x+\left(1+2+3+...+100\right)=15150\)

\(2x+\left[\left(1+100\right)+\left(2+99\right)+...+\left(50+51\right)\right]=15150\)

\(2x+\left[101+101+...+101\right]=15150\)CÓ 50 SỐ 101

\(2x+\left[101\times50\right]=15150\)

\(2x=15150:5050\)

\(2x=3\)

\(x=3:2\)

\(x=1.5\)

a, 2x + (1+2+3+4+...+100) = 15150 

=> 2x + \(\frac{\left(1+100\right).\left[\left(100-1\right)+1\right]}{2}\)= 15150 

=> 2x + \(\frac{101.100}{2}\)= 15150 

=> 2x + 5050 = 15150 

=> 2x             = 15150 - 5050 

=> 2x             = 10100

=> x              =  10100 : 2 

=> x              = 5050 

Vậy x = 5050 

b, .(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8)=36 

=> (x + x + x + x +x + x +x +x ) + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) = 36 

=> 8x + 36 = 36 

=> 8x         = 0 

=>  x          = 0 

Vậy x = 0 

c, 0+0+4+6+8+...+2x=110 

Sửa đề :0 + 2 + 4 + 6 + 8 + ... + 2x = 110 = 2 + 4 + 6 + 8 + ... + 2x = 110 

SSH  : \(\frac{\left(2\text{x}-2\right)}{2}+1=x-1+1=x\)

Tổng : \(\frac{\left(2\text{x}+2\right).x}{2}=110\Leftrightarrow\frac{2.\left(x+1\right).x}{2}=110\)

                                                    \(\Leftrightarrow\left(x+1\right)x=110\)

                                                     \(\Leftrightarrow\left(10+1\right).10=110\)

 => x = 10 

Vậy x = 10 

Dấu ^ là nhân hay là mũ vậy bạ?

Áp dụng BĐT Cauchy-Schwarz ta có:

\(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}=\frac{6^2}{3}=12\)

Dấu " = " xảy ra <=> a=b=c=2

ĐK  \(x\ge0\)

Đặt \(x=a,x+1=b\)

\(PT\Leftrightarrow a^4+b^4=\left(a+b\right)^4\)

<=> 4a³b+6a²b²+4ab³=0

<=> ab(2a²+3ab+2b²)=0

=>ab=0 (vì 2a²+3ab+2b²>0)

=>\(\orbr{\begin{cases}a=0\\b=0\end{cases}}\)<=>\(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy.............................

#)Mình trả lời nhanh luôn nhé :

Khi nhân số 142 857 với 2,3,4,5,6 sẽ thấy : tích của chúng đều được viết bởi 6 số 1,4,2,8,5,7

Ai chẳng biết

Ta có \(a^2+b^2+c^2\ge ab+bc+ac\)

Áp dụng 

=> \(a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\ge a^2bc+ab^2c+abc^2=abc\left(a+b+c\right)\)

=> \(\frac{1}{a^4+b^4+c^4+abcd}\le\frac{1}{abc\left(a+b+c+d\right)}\)

Khi đó 

\(VT\le\frac{1}{a+b+c+d}\left(\frac{1}{abc}+\frac{1}{bcd}+\frac{1}{cda}+\frac{1}{dab}\right)\)

=> \(VT\le\frac{1}{a+b+c+d}.\frac{a+b+c+d}{abcd}=1\)

Dấu bằng xảy ra khi \(a=b=c=d=1\)

Vậy MaxA=1 khi a=b=c=d=1

a;b;c la so thuc thi chua chac a;b;c > 0 dau

\(\frac{x}{6}=\frac{1}{y}=\frac{1}{2}\)

\(\frac{x}{6}=\frac{1}{2}\Rightarrow6.1=2.x=6:2\)

\(\Rightarrow x=3\)

\(\frac{1}{y}=\frac{1}{2}\Rightarrow2.1=1.y=2:1\)

\(\Rightarrow y=2\)

\(\Leftrightarrow\frac{3}{6}=\frac{1}{2}=\frac{1}{2};x=3;y=2\)

#)Giải :

Ta có : \(\frac{x}{6}=\frac{1}{y}\Rightarrow xy=6.1\)

\(\Rightarrow xy=6\)

\(\Rightarrow\orbr{\begin{cases}x=1\Rightarrow y=6\\x=2\Rightarrow y=3\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=3\Rightarrow y=2\\x=6\Rightarrow y=1\end{cases}}\)