a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

1. \(\frac{-17}{21}:\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)

\(-\frac{17}{21}:\frac{17}{20}< x+\frac{4}{7}< \frac{7}{12}\)

\(-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)

\(-\frac{80}{84}< \frac{84x+48}{84}< \frac{49}{84}\)

\(-80< 84x+48< 49\)

\(\begin{cases}-80< 84x+48\\84x+48< 49\end{cases}\) 

\(\begin{cases}84x>-128\\84x< 1\end{cases}\)

\(\begin{cases}x>-\frac{32}{21}\\x< \frac{1}{84}\end{cases}\)

\(\Rightarrow-\frac{32}{21}< x< \frac{1}{84}\)

\(-\frac{17}{21}\div\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)

\(-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)

\(-\frac{32}{21}< x< \frac{1}{84}\)

\(-1^{11}_{21}< x< \frac{1}{84}\)

\(\Rightarrow x\in\left\{-1;0\right\}\)

Vậy x = 0

\(\frac{4}{3}\times1,25\times\left(\frac{16}{5}-\frac{5}{16}\right)< 2x< 4-\frac{4}{3}+3-\frac{3}{2}+2\)

\(\frac{77}{16}< 2x< \frac{37}{6}\)

\(\frac{77}{32}< x< \frac{37}{12}\)

\(2^{13}_{32}< x< 3^1_{12}\)

=> x = 3

2. \(\frac{4}{3}.1,25\left(\frac{16}{5}-\frac{5}{16}\right)< 2x< 4-\frac{4}{3}+3-\frac{3}{2}+2\)

\(\frac{16}{15}.\frac{231}{80}< 2x< \frac{37}{6}\)

\(\frac{77}{25}< 2x< \frac{37}{6}\)

\(\begin{cases}\frac{77}{25}< 2x\\2x< \frac{37}{6}\end{cases}\)

\(\begin{cases}2x>\frac{77}{25}\\x< \frac{37}{12}\end{cases}\)

\(\begin{cases}x>\frac{77}{50}\\x< \frac{37}{12}\end{cases}\)

\(\frac{77}{50}< x< \frac{37}{12}\)

vì x+y=4 nền (x+y)^2=4^2                                                                                                                                                                                            =x^2+ 2xy+y^2=16        ma  xy=5 nên 2xy=10  ta có x^2+y^2+10=16 ; x^2+y^2= 16-10                                                                                                                                                                                     x^2+y^2=6                                     kết quả mik là z đó nhưng k biết có đúng k bn ak

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

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

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

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

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

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}

a) \(6xy+4x-9y-7=0\)

  \(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)

\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)

\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)

Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)

Tự làm típ

\(A=x^3+y^3+xy\)

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

\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))

\(A=x^2+y^2\)

Áp dụng bất đẳng thức Bunhiakovxky ta có :

\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)

\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)

\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)

Hay \(x^3+y^3+xy\ge\frac{1}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)

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2.2².2³.2⁴....2^x=32768

\(\Leftrightarrow2^{1+2+...+x}=2^{15}\)

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

Đặt \(A=1+2+...+x\)

Tổng A có số số hạng là:

\(\left(x-1\right):1+1=x\)(số)

Tổng A theo x là:

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

THay vào ta có:

\(\frac{x^2+x}{2}=15\Leftrightarrow x^2+x=30\)

\(\Leftrightarrow x^2+6x-5x-30=0\)

\(\Leftrightarrow x\left(x+6\right)-5\left(x+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=5\left(tm\right)\\x=-6\left(loai\right)\end{array}\right.\)