\(\frac{3}{5}x+\frac{2}{3}=\frac{4}{5}\)

\(\Leftrightarrow\frac{3}{5}x=\frac{4}{5}-\frac{2}{3}=\frac{2}{15}\)

\(\Leftrightarrow x=\frac{2}{15}:\frac{3}{5}=\frac{10}{45}=\frac{2}{9}\)

\(\left|\frac{1}{2}x+\frac{3}{5}\right|=\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x+\frac{3}{5}=\frac{1}{2}\\\frac{1}{2}x+\frac{3}{5}=-\frac{1}{2}\end{cases}}\)

\(-\frac{625}{256}\)nha

\(\frac{-5^4}{4^4}\)=\(\frac{-625}{256}\)

A D B C a b 1 1 1 M N 1 1 c 2 2 2

a) Ta có \(\widehat{B_1}+\widehat{ABC}=180^o\)( 2 góc kề bù )

\(\Rightarrow70^o+\widehat{ABC}=180^o\)\(\Leftrightarrow\widehat{ABC}=110^o\)mà \(\widehat{C_1}=110^o\)

\(\Rightarrow\widehat{ABC}=\widehat{C_1}\)mà \(\widehat{ABC}\text{ và }\widehat{C_1}\)là 2 góc đồng vị

=> a // b

b) Vì \(\hept{\begin{cases}a//b\left(\text{ phần a}\right)\\c⊥b\end{cases}}\Rightarrow c⊥a\Rightarrow\widehat{A_1}=90^o\)

c) Ta có : \(\widehat{B_1}=\widehat{C_2}\)( 2 góc so le trong )\(\Rightarrow\widehat{C_2}=70^o\)

Lại có tia phân giác của \(\widehat{ABC}\)tạo ra \(\widehat{B_2}=\frac{\widehat{ABC}}{2}=\frac{110^o}{2}=55^o\)

Mà \(\widehat{N_2}\)là góc ngoài của ∆NBC \(\Rightarrow\widehat{N_2}=\widehat{B_2}+\widehat{C_2}=70^o+55^o=125^o\)

Lại có : \(\hept{\begin{cases}\widehat{N_1}+\widehat{N_2}=180^o\left(\text{2 góc kề bù }\right)\\\widehat{N_2}=\widehat{D_2}+\widehat{M_1}\left(\text{do }\widehat{N_2}\text{ là góc ngoài của }∆DMN\right)\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\widehat{N_1}+125^o=180^o\\\widehat{M_1}=\widehat{N_2}-\widehat{D_2}\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\widehat{N_1}=180^o-125^o\\\widehat{M_1}=125^o-90^o\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\widehat{N_1}=55^o\\\widehat{M_1}=35^o\end{cases}}\)

@ln2a2_lemanhhung

cop bài thì đọc cho kĩ đề vô nhé bạn 

Ta có :

\(\frac{x}{3}=\frac{y}{4}\)\(\Rightarrow\left(\frac{x}{3}\right)^2=\left(\frac{y}{4}\right)^2=\frac{x^2}{3^2}=\frac{y^2}{4^2}=\frac{x}{3}.\frac{y}{4}=\frac{xy}{3.4}=\frac{12}{12}=1=\left(\pm1\right)^2\)

\(\Rightarrow\hept{\begin{cases}x^2=\left(±1\right)^2.3^2\\y^2=\left(±1\right)^2.4^2\end{cases}}\Rightarrow\hept{\begin{cases}x^2=\left[\left(\pm1\right).3\right]^2\\y^2=\left[\left(\pm1\right).4\right]^2\end{cases}}\Rightarrow\hept{\begin{cases}x^2=\left(\pm3\right)^2\\y^2=\left(\pm4\right)^2\end{cases}}\Rightarrow\hept{\begin{cases}x=\pm3\\y=\pm4\end{cases}}\)

Mà x và y cùng dấu => ( x , y ) ∈ { ( -3 , -4 ) ; ( 3 , 4 ) }

Cần j gì vậy bn

ta có: 

1C

2C

3D

4B

1 c

2c

3d

4b

câu hỏi đúng 100%

ab200

\(\left|x\right|\ge0\)

\(\Rightarrow\frac{-1}{\left|x\right|+1}\ge\frac{-1}{0+1}=-1\)

Dấu bằng khi x = 0

1 + 1 = 2

@Cỏ

#Forever

1+1=2 nha bạn!

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