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

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

\(=2^3.3=24\)

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

\(\dfrac{9^{15}.8^{11}}{3^{29}.16^8}=\dfrac{3^{30}.2^{33}}{3^{29}.2^{32}}=3.2=6\)

\(\dfrac{2^{11}.9^3}{3^5.16^2}=\dfrac{2^{11}.3^6}{3^5.2^8}=2^3.3=8.3=24\)

Câu 1:
TXĐ:D=R

\(f\left(-x\right)=2\cdot\left(-x\right)^4-3\cdot\left(-x\right)^2+1\)

\(=2x^4-3x^2+1=f\left(x\right)\)

=>f(x) là hàm số chẵn

\(a,\Rightarrow x-2=8\\ \Rightarrow x=10\\ b,\Rightarrow x+12-17=20\\ \Rightarrow x-5=20\\ \Rightarrow x=25\\ c,\Rightarrow11-\left(4x+5\right):3=4\\ \Rightarrow\left(4x+5\right):3=7\\ \Rightarrow4x+5=21\\ \Rightarrow x=4\\ d,\Rightarrow\left(35:x+3\right)\cdot17=136\\ \Rightarrow35:x+3=8\\ \Rightarrow35:x=5\\ \Rightarrow x=7\\ e,\Rightarrow41-\left(2x-5\right)=720:8\cdot5=180\\ \Rightarrow2x-5=-139\\ \Rightarrow2x=-134\\ \Rightarrow x=-67\)

\(2,\\ a,\Rightarrow x^2=4^3:16=64:16=4=2^2=\left(-2\right)^2\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ b,\Rightarrow\left(x-1\right)^2=9=3^2=\left(-3\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\\ c,\Rightarrow\left(3x-7\right)^5=2^5\\ \Rightarrow3x-7=2\\ \Rightarrow3x=9\Rightarrow x=3\)

a: góc ASB=1/2*180=90 độ=góc ABM

b: ON vuông góc AS

BS vuông góc SA

=>ON//BS

c: góc OIM+góc OBM=180 độ

=>OIMB nội tiếp

Bài 1.2

1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

2) Ta có: \(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}+1}{3-\sqrt{x}}-\dfrac{3-11\sqrt{x}}{x-9}\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)