\(\frac{2}{9}-6=\frac{2}{9}-\frac{54}{9}=-\frac{52}{9}\)

Lớp 4 chưa học số âm mà.

trả lời hộ cái

2/9-6=52 /9

\(\dfrac{4}{x+3}=\dfrac{2}{x-1}\left(x\ne-3;x\ne1\right)\)

suy ra: \(4\left(x-1\right)=2\left(x+3\right)\\ < =>4x-4=2x+6\\ < =>4x-2x=6+4\\ < =>2x=10\\ < =>x=5\left(tm\right)\)

\(\dfrac{4}{x+3}=\dfrac{2}{x-1}\text{ĐKXĐ}:x\ne-3;1\)

\(\Leftrightarrow\dfrac{4\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-1\right)}MTC:\left(x+3\right)\left(x-1\right)\)

\(\Rightarrow4x-4=2x+6\)

\(\Leftrightarrow4x-4-2x-6=0\)

\(\Leftrightarrow2x-10=0\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\left(\text{nhận}\right)\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{5\right\}\)

ĐKXĐ: `x+3\ne0=>x\ne-3` ; `x-1\ne0=>x\ne1`

\(\dfrac{4}{x+3}=\dfrac{2}{x-1}\\ \Leftrightarrow4\left(x-1\right)=2\left(x+3\right)\\ \Leftrightarrow4x-4=2x+6\\ \Leftrightarrow4x-2x=6+4\\ \Leftrightarrow2x=10\\ \Leftrightarrow x=5\left(tm\right)\)

\(\left(\dfrac{2}{3}\right)^2.\left(\dfrac{2}{3}\right)^3\)

\(=\left(\dfrac{2}{3}\right)^{2+3}\)

\(=\left(\dfrac{2}{3}\right)^5\)

Chúc bạn học tốt!

vậy hông 

\(\frac{2^2}{2^{x+y}}=8\)và \(\frac{3^2.\left(x+y\right)}{3^{5y}}=343\)

a) Ta có: \(\left(2x+3\right)^2-4\left(x-2\right)\left(x+2\right)\)

\(=4x^2+12x+9-4\left(x^2-4\right)\)

\(=4x^2+12x+9-4x^2+16\)

\(=12x+25\)

b) Ta có: \(\dfrac{x+6}{x^2-4}-\dfrac{2}{x\left(x+2\right)}\)

\(=\dfrac{x\left(x+6\right)}{x\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x^2+6x-2x+4}{x\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x^2+4x+4}{x\left(x+2\right)\left(x-2\right)}\)

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

\(=\dfrac{x+2}{x\left(x-2\right)}\)

\(\dfrac{6x-2}{x+4}=5\Leftrightarrow6x-2=5x+20\)

\(\Leftrightarrow x=22\)

\(\dfrac{6x-2}{x+4}=5\)

ĐKXĐ:  \(x\ne-4\)

\(\dfrac{6x-2}{x+4}=\dfrac{5\left(x+4\right)}{x+4}\)

\(6x-2=5x+20\)

\(6x-5x=20+2\)

\(x=22\)

Vậy \(S=\left\{22\right\}\)

\(\dfrac{6x-2}{x+4}=5\left(dkxd:x\ne-4\right)\)

Suy ra :

\(6x-2=5\left(x+4\right)\)

\(\Rightarrow6x-2=5x+20\)

\(\Rightarrow x=22\left(tmdk\right)\)

Vậy \(S=\left\{22\right\}\)

\(\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{10}{-x^2+4}\)

\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-10\)

\(\Leftrightarrow x^2+4x+4-x^2+4x-4=-10\)

=>8x=-10

hay x=-5/4

\(y=\dfrac{x^2-3x+2}{x-1}=\dfrac{(x-1)(x-2)}{x-1}=x-2\\ \Rightarrow y'=1\)

\(\dfrac{x+2}{-4}=-\dfrac{9}{x+2}\\ \Rightarrow\left(x+2\right)^2=\left(-4\right).\left(-9\right)\\ \Rightarrow\left(x+2\right)^2=36\\ \Rightarrow\left(x+2\right)^2=\pm6^2\\ \Rightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)

=>(x+2)^2=36

=>x+2=6 hoặc x+2=-6

=>x=4 hoặc x=-8