1. \(x^{10}=25x^8\Leftrightarrow x^{10}:x^8=25\Leftrightarrow x^2=25=5^2\Leftrightarrow x=5\)

2. \(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=\frac{2^{40}}{2^{30}}=2^{10}\)

1)\(x^{10}=25x^8\)

\(\Rightarrow x^{10}:x^8=25\)

\(\Rightarrow x^2=5^2\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

2)\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}\)

A x y C B 60độ 30độ 60độ 30độ

Vì \(\widehat{xAB}\) và \(\widehat{ABC}\) là hai góc so le trong bằng nhau (= 60^o)

=> Ax và BC song song với nhau (1)

Vì \(\widehat{yAC}\)và \(\widehat{ACB}\)là hai góc so le trong bằng nhau (= 30^o)

=> Ay và BC song song với nhau (2)

Từ (1) và (2) => BC // Ax // Ay

                   hay BC // xy    (ĐCCM)

2x + 3z = 2y + 118 => 2x + 3z - 2y = 118 

Theo bài ra: 3x = 4y ; 5y = 7z 

=> 15x = 20y = 28 z 

=> \(\frac{x}{\frac{1}{15}}=\frac{y}{\frac{1}{20}}=\frac{z}{\frac{1}{28}}=\frac{2x-2y+3z}{\frac{2}{15}-\frac{2}{20}+\frac{3}{28}}=\frac{118}{\frac{59}{420}}=840\)

=> x = 56; y = 42; z = 30

4x - 3y = 0 <=> 4x = 3y => 20 x = 15y ; 5y = 3z => 15y = 9z 

=> 20 x = 15y = 9z 

=> \(\frac{x}{\frac{1}{20}}=\frac{y}{\frac{1}{15}}=\frac{z}{\frac{1}{9}}=\frac{3x-2y+z}{\frac{3}{20}-\frac{2}{15}+\frac{1}{9}}=\frac{46}{\frac{23}{180}}=360\)

=>  x = 18; y = 24; z= 40

Với \(y^2=zx;z^2=xy\)và ĐK : \(x+y-z=1\), ta có : \(y\cdot y=z\cdot x;z\cdot z=x\cdot y\)và ĐK : \(x+y-z-1=0\).

Với \(x+y-z-1=0\), coi \(1=a\), và chỉ khi \(x+y-z=a\)thì \(x+y-z-a=0\)( vì \(a=1\))

\(x+y-z-a=0\Rightarrow x+y-\left(z+a\right)\Rightarrow x+y=z+a\)(ĐK : \(y^2=zx;z^2=xy;x+y-z=a\))

Vậy thỏa mãn \(x=y=z=1\).

\(2^{x-3}-3.2^x=-92\)

\(2^x:2^3-3.2^x=-92\)

\(2^x.\left(\frac{1}{8}-3\right)=-92\)

\(2^x.\frac{-23}{8}=-92\)

\(2^x=32\)

\(2^x=2^5\)

\(\Rightarrow x=5\)

GIÚP VỚI

HELP ME HELP ME !!!

a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)

b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)

c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)

\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)

\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)