Thi vòng 12 à bạn!!! Để mk chép đề mà làm 

ta có \(x^2+\frac{1}{x^2}\)

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

=> \(\left(x+\frac{1}{x}\right)^2=25.vì\)\(x>0\Rightarrow x+\frac{1}{x}>0\Rightarrow x+\frac{1}{x}=5\)

\(\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3x+\frac{3}{x}=x^3+\frac{1}{x^3}+15\)

\(\Rightarrow x^3+\frac{1}{x^3}=5^3+15=110\)

\(\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=x^5+\frac{1}{x^5}+x+\frac{1}{x}=x^5+\frac{1}{x^5}+5\)

\(\Rightarrow x^5+\frac{1}{x^5}=23\cdot110-5=2525\)

Vậy...

\(2\cdot2^2\cdot2^3\cdot2^4\cdot\cdot\cdot2^x=32768\)

\(\Leftrightarrow2^{1+2+3+4+\cdot\cdot\cdot+x}=2^{15}\)

\(\Leftrightarrow1+2+3+4+..+x=15\)

\(\Leftrightarrow\)\(\frac{\left(1+x\right)x}{2}=15\)

\(\Leftrightarrow x\left(x+1\right)=30=5\left(5+1\right)\)

Vậy x=5

Bài 2:

Bậc của đơn thức là 2+5+3=10

Bài 3:

\(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\)

\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=5\)

+)TH1: \(x\ge\frac{1}{4}\) thì bt trở thành

\(2x-\frac{1}{2}=5\Leftrightarrow2x=\frac{11}{2}\Leftrightarrow x=\frac{11}{4}\left(tm\right)\)

+)TH2: \(x< \frac{1}{4}\) thì pt trở thành

\(2x-\frac{1}{2}=-5\Leftrightarrow2x=-\frac{9}{2}\Leftrightarrow x=-\frac{9}{4}\left(tm\right)\)

Vậy x={-9/4;11/4}

Bài 2:

TH1: \(x\le-\frac{5}{2}\)

<=>\(-\left(x+\frac{5}{2}\right)+\frac{2}{5}-x=0\)<=>\(-x-\frac{5}{2}+\frac{2}{5}-x=0\)<=>\(-\frac{21}{10}-2x=0\)

<=>\(-2x=\frac{21}{10}\)<=>\(x=\frac{-21}{20}\)(loại)

TH2: \(-\frac{5}{2}< x\le\frac{2}{5}\)

<=>\(x+\frac{5}{2}+\frac{2}{5}-x=0\)<=>\(\frac{29}{10}=0\)(loại)

TH3: \(x>\frac{2}{5}\)

<=>\(x+\frac{5}{2}+x-\frac{2}{5}=0\)<=>\(2x+\frac{21}{10}=0\)<=>\(2x=-\frac{21}{10}\)<=>\(x=-\frac{21}{20}\)(loại)

Vậy không có số x thỏa mãn đề bài

Bài 1:

Vì \(\left(x-2\right)^2\ge0\) nên\(\left(x-2\right)^2\le0\) khi \(\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)

Bài 3:

Đặt \(\frac{x}{15}=\frac{y}{9}=k\Rightarrow\hept{\begin{cases}x=15k\\y=9k\end{cases}}\)

Theo đề bài: xy=15 <=> 15k.9k=135k²=15 <=> k²=1/9 <=> k=-1/3 hoặc k=1/3

+) \(k=-\frac{1}{3}\Rightarrow\hept{\begin{cases}x=\left(-\frac{1}{3}\right).15=-5\\y=\left(-\frac{1}{3}\right).9=-3\end{cases}}\)

+) \(k=\frac{1}{3}\Rightarrow\hept{\begin{cases}x=\frac{1}{3}.15=5\\y=\frac{1}{3}.9=3\end{cases}}\)

Vậy ...........

2) \(ĐKXĐ:x\notin\left\{-2;-3;-4\right\}\)

PT <=> \(x+\frac{x}{x+2}+\frac{x+3}{x^2+3x+2x+6}+\frac{x+4}{x^2+4x+2x+8}-1=0\)

<=>\(x+\frac{x}{x+2}+\frac{x+3}{x\left(x+3\right)+2\left(x+3\right)}+\frac{x+4}{x\left(x+4\right)+2\left(x+4\right)}-1=0\)

<=>\(x+\frac{x}{x+2}+\frac{x+3}{\left(x+2\right)\left(x+3\right)}+\frac{x+4}{\left(x+2\right)\left(x+4\right)}-1=0\)

<=>\(x+\frac{x}{x+2}+\frac{1}{x+2}+\frac{1}{x+2}-1=0\)

<=>\(x+\frac{x+1+1}{x+2}-1=0\)

<=>\(x+\frac{x+2}{x+2}-1=0\Leftrightarrow x+1-1=0\Leftrightarrow x=0\)

Vậy x=0 thì thỏa mãn PT

2/ \(\frac{1}{2}x2y5z3=\left(\frac{1}{2}.2.5.3\right)xyz\)\(=15xyz\)

\(\Rightarrow\frac{1}{2}x2y5z3\)có bậc là 3

3/ \(\frac{x}{4}=\frac{9}{x}\Leftrightarrow x^2=9.4\Rightarrow x^2=36\) mà \(x>0\Rightarrow x=6\)

4/ \(\left|2x-\frac{1}{2}\right|+\frac{3}{7}=\frac{38}{7}\Rightarrow\left|2x+\frac{1}{2}\right|=\frac{35}{7}=5\Rightarrow\hept{\begin{cases}2x+\frac{1}{2}=5\Rightarrow2x=\frac{9}{2}\Rightarrow x=\frac{9}{4}\\2x+\frac{1}{2}=-5\Rightarrow2x=\frac{-11}{2}\Rightarrow x=\frac{-11}{4}\end{cases}}\)

Câu hỏi của cai j vay - Toán lớp 8 - Học toán với OnlineMath

Em tham khảo nhé!

Lớp 8 thì

Hôm nay thi cấp huyện mà

Fải k?//

Thi tốt nghen>>>~~~~

\(A=\frac{6x^2+8x+7}{x^3-1}+\frac{x}{x^2+x+1}+\frac{6}{1-x}\)

<=>\(A=\frac{6x^2+8x+7}{x^3-1}+\frac{\left(x-1\right)x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{\left(-6\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)

<=>\(A=\frac{6x^2+8x+7}{x^3-1}+\frac{x^2-x}{x^3-1}+\frac{-6x^2-6x-6}{x^3-1}\)

<=>\(A=\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)<=>\(A=\frac{1}{x-1}\)<=>\(4A=\frac{4}{x-1}\)

Theo đề bài 4A=x-1 => \(4A=\frac{4}{x-1}=x-1\Rightarrow\left(x-1\right)^2=4\Rightarrow\orbr{\begin{cases}x-1=-2\\x-1=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Vì x<0 nên x=-1

thay z = -(x+y) , y = -(z+x),... vao

=> Duoc bieu thuc trong do co 1/xy + 1/yz + 1/zx = (x+y+z)/xyz = 0