Ta có: \(3x=7y\)

\(\Rightarrow\frac{x}{7}=\frac{y}{3}=\frac{y-x}{3-7}=\frac{8}{-4}=-2\)

\(\Rightarrow\hept{\begin{cases}x=-14\\y=-6\end{cases}}\)

\(\frac{x+2}{3}=\frac{y-1}{4}=\frac{z+3}{5}\)

\(=\frac{x+2-y+1+z+3}{3-4+5}\)

\(=\frac{x-y+z+6}{4}=\frac{66}{4}=16,5\)

\(\Rightarrow x,y,z=....\)

\(3^{x-1}+\frac{5}{3}.3^x=162\)

\(\Leftrightarrow3^x.\frac{1}{3}+\frac{5}{3}.3^x=162\)

\(\Leftrightarrow3^x.2=162\)

\(\Leftrightarrow3^x=81\Leftrightarrow x=4\)

Bài 1: Tính hợp lí:

\(a)\)\(\frac{14}{57}+\frac{29}{23}-\frac{71}{57}+\frac{-6}{23}\)

\(=\)\(\left(\frac{14}{57}-\frac{71}{57}\right)+\left(\frac{29}{23}+\frac{-6}{23}\right)\)

\(=\)\(\left(-\frac{57}{57}\right)+\left(\frac{23}{23}\right)\)

\(=\)\(\left(-1\right)+1\)

\(=\)\(0\)

\(b)\)\(\frac{5}{12}.\left(\frac{-3}{4}\right)+\frac{7}{12}.\left(\frac{-3}{4}\right)\)

\(=\)\(\left(\frac{-3}{4}\right).\left(\frac{5}{12}+\frac{7}{12}\right)\)

\(=\)\(\left(\frac{-3}{4}\right).1\)

\(=\)\(-\frac{3}{4}\)

\(c)\)\(\left(\frac{-3}{11}\div\frac{5}{22}\right).\left(\frac{-15}{3}\div\frac{26}{3}\right)\)

\(=\)\(\left(\frac{-3}{11}.\frac{22}{5}\right).\left(\frac{-15}{3}.\frac{3}{26}\right)\)

\(=\)\(\frac{-6}{5}.\frac{-15}{26}\)

\(=\)\(\frac{9}{13}\)

\(d)\)\(\left(0,25\right)^{100}.4^{103}\)

\(=\)\(\left(\frac{1}{4}\right)^{100}.4^{103}\)

\(=\)\(\left(\frac{1}{4}.4\right)^{100+103}\)

\(=\)\(1^{203}\)

\(=\)\(1\)

Bài 2: Tìm x, biết:

\(a)\)\(\left(\frac{3}{7}-2x\right)^2=\frac{4}{9}\)

\(\Leftrightarrow\)\(\left(\frac{3}{7}-2x\right)^2=\left(\frac{2}{3}\right)^2\)

\(\Leftrightarrow\)\(\frac{3}{7}-2x=\frac{2}{3}\)

\(\Leftrightarrow\)\(2x=\frac{3}{7}-\frac{2}{3}\)

\(\Leftrightarrow\)\(2x=\frac{-5}{21}\)

\(\Leftrightarrow\)\(x=\frac{-5}{21}\div2\)

\(\Leftrightarrow\)\(x=\frac{-5}{42}\)

\(b)\)\(x+5,5=7,5\)

\(\Leftrightarrow\)\(x=7,5-5,5\)

\(\Leftrightarrow\)\(x=2\)

TL:

5^x+2. 4^y = 50^x

= 5³ . 4¹ = 50¹

=  ( 5³ . 4¹ ) : 50

= 6550 : 50

= 131

TL

5^x+2. 4^y = 50^x

= 131

HT

7,923~7,92

17,418~17,42

79,1364~79,14

50,401~50,40

0,155~0,16

60,996~61,00

giải đi tặng 50 sao

Hướng dẫn thoai , bìa này tư duy đc :))

a) các cặp song song là : QE//PH;PQ//FH

b)

Góc EOQ = 90^o - góc OQE = 90^o- 55^o = 35^o

Góc OFH = 90^o - góc EOQ = 90^o - 35^o = 55^o

Góc PHF = 90^o - góc OFH = 90^o - 55^o = 35^o

\(\frac{x}{z}=\frac{z}{y}\Rightarrow z^2=xy\)

+) \(\frac{x^2+z^2}{y^2+z^2}=\frac{x^2+xy}{y^2+xy}=\frac{x\left(x+y\right)}{y\left(x+y\right)}=\frac{x}{y}\left(\text{Đ}pcm\right)\)

Học tốt =))

@OneChampCap : Không phải bài nào dạng này cũng làm như vậy được :>>

Giải :

Ta có : \(\frac{x}{z}=\frac{z}{y}\Rightarrow\left(\frac{x}{z}\right)^2=\frac{x^2}{z^2}=\left(\frac{z}{y}\right)^2=\frac{z^2}{y^2}=\frac{x}{z}.\frac{z}{y}=\frac{x}{y}\)

Áp dụng tính chất dãy tỉ số bằng nhau , ta có :

\(\left(\frac{x}{z}\right)^2=\frac{x^2}{z^2}=\left(\frac{z}{y}\right)^2=\frac{z^2}{y^2}=\frac{x}{z}.\frac{z}{y}=\frac{x}{y}=\frac{x^2+z^2}{y^2+z^2}\)

Vậy \(\frac{x}{y}=\frac{x^2+z^2}{y^2+z^2}\)