\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}\)

Ta lấy vễ trên chia vế dưới

\(=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}\)

Ta lấy vế trên chia vế dưới

\(=2^3.3=24\)

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{\left(3^2\right)^{15}.\left(2^3\right)^{11}}{3^{29}.\left(2^4\right)^8}=\dfrac{3^{30}.2^{33}}{3^{29}.3^{32}}=3.2=6\)
\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.\left(3^2\right)^3}{3^5.\left(2^4\right)^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

Bài 1:

a: \(-\frac52-\frac49\cdot\frac{21}{16}\)

\(=-\frac52-\frac{4}{16}\cdot\frac{21}{9}\)

\(=-\frac52-\frac14\cdot\frac73=-\frac52-\frac{7}{12}=\frac{-30-7}{12}=-\frac{37}{12}\)

b: \(\frac{-5}{4}\cdot\frac{-3}{7}+\frac{-5}{4}\cdot\frac{17}{7}+12\frac12\)

\(=-\frac54\left(-\frac37+\frac{17}{7}\right)+12,5\)

\(=-\frac54\cdot\frac{14}{7}+12,5=-\frac54\cdot2+12,5=-\frac52+12,5=10\)

c: \(\left(-\frac13\right)^2+60\%+2\frac89-\frac85\)

\(=\frac19+\frac35+2+\frac89-\frac85\)

\(=2+\left(\frac19+\frac89\right)+\left(\frac35-\frac85\right)=2\)

d: \(15\frac37-\left(\frac89+7\frac37\right)\)

\(=15+\frac37-\frac89-7-\frac37\)

\(=8-\frac89=\frac{72-8}{9}=\frac{64}{9}\)

Bài 2:

a: \(1,75-20\%\cdot x=\frac15\)

=>\(0,2x=1,75-0,2=1,55\)

=>\(x=1,55:0,2=1,55\cdot5=7,75\)

b: \(\left(\frac23-x\right):\frac{-3}{10}=75\%\)

=>\(\frac23-x=\frac34\cdot\frac{-3}{10}=\frac{-9}{40}\)

=>\(x=\frac23+\frac{9}{40}=\frac{80+27}{120}=\frac{107}{120}\)

c: \(\left|\frac34x-\frac15\right|=0,25\)

=>\(\left[\begin{array}{l}\frac34x-\frac15=\frac14\\ \frac34x-\frac15=-\frac14\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac34x=\frac15+\frac14=\frac{9}{20}\\ \frac34x=-\frac14+\frac15=\frac{-1}{20}\end{array}\right.\)

=>\(\left[\begin{array}{l}x=\frac{9}{20}:\frac34=\frac{9}{20}\cdot\frac43=\frac{36}{60}=\frac35\\ x=-\frac{1}{20}:\frac34=-\frac{1}{20}\cdot\frac43=\frac{-4}{60}=-\frac{1}{15}\end{array}\right.\)

d: \(x-\frac29x=\frac59\)

=>\(x\left(1-\frac29\right)=\frac59\)

=>\(x\cdot\frac79=\frac59\)

=>\(x=\frac59:\frac79=\frac57\)

jimmmmmmmmmmmmmmmmmmmmmmmmmmm

he he he he he he

Bài 3:

a: Hệ số tỉ lệ a là: \(a=7\cdot10=70\)

b: xy=70

=>\(y=\frac{70}{x}\)

c: Khi x=5 thì \(y=\frac{70}{5}=14\)

Khi x=14 thì \(y=\frac{70}{14}=5\)

d: Khi y=15 thì \(x=\frac{70}{y}=\frac{70}{15}=\frac{14}{3}\)

Khi y=3 thì \(x=\frac{70}{y}=\frac{70}{3}\)

Bài 4:

a: Xét ΔMNI và ΔMPI có

MN=MP

NI=PI

MI chung

Do đó: ΔMNI=ΔMPI

b: ΔMNI=ΔMPI

=>\(\hat{MNI}=\hat{MPI}\)

c: ΔMNI=ΔMPI

=>\(\hat{NMI}=\hat{PMI}\)

=>MI là phân giác của góc NMP

d: ΔMNI=ΔMPI

=>\(\hat{MIN}=\hat{MIP}\)

mà \(\hat{MIN}+\hat{MIP}=180^0\) (hai góc kề bù)

nên \(\hat{MIN}=\hat{MIP}=\frac{180^0}{2}=90^0\)

=>MI⊥NP

mà I là trung điểm của NP

nên MI là đường trung trực của NP

Bài 3:

a. \(R=R1+R2=15+30=45\Omega\)

b. \(\left\{{}\begin{matrix}I=U:R=9:45=0,2A\\I=I1=I2=0,2A\left(R1ntR2\right)\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}U1=R1.I1=15.0,2=3V\\U2=R2.I2=30.0,2=6V\end{matrix}\right.\)

Bài 4:

\(I1=U1:R1=6:3=2A\)

\(\Rightarrow I=I1=I2=2A\left(R1ntR2\right)\)

\(U=R.I=\left(3+15\right).2=36V\)

\(U2=R2.I2=15.2=30V\)

(3x)^2=3^2.x^2=9x^2