Đề sai

Ta có : \(\hept{\begin{cases}a+3b=8\\2a+3c=7\end{cases}}\Rightarrow\left(a+3b\right)+\left(2a+3c\right)=8+7\)

\(\Leftrightarrow a+3b+2a+3c=15\)

\(\Leftrightarrow\left(2a+a\right)+3b+3c=15\)

\(\Leftrightarrow3a+3b+3c=15\)

\(\Leftrightarrow3\left(a+b+c\right)=15\)

\(\Leftrightarrow a+b+c=15\div3\)

\(\Leftrightarrow a+b+c=5\)

Đề đúng đấy ạ :)))

A=124-5(x-7)

A= 159-5x

Để A có giá trị lớn nhất thì -5x=0\(\Rightarrow\) x=0

Vậy \(A_{max}\) =159 khi \(x\)=0

\(3^{-1}.3^x+5.3^{x-1}=162\)

=> \(\frac{1}{3}.3^x+5.3^x:3=162\)

=> \(\left(\frac{1}{3}+\frac{5}{3}\right).3^x=162\)

=> 2 . 3^x = 162

=> 3^x = 162 : 2

=> 3^x = 81

=> 3^x = 3⁴

=> x  = 4

Vậy ...

\(3^{-1}.3^x+5.3^{x-1}=162\)

\(\Rightarrow3^{-1+x}+5.3^{x-1}=162\)

\(\Rightarrow3^{x-1}+5.3^{x-1}=162\)

\(\Rightarrow3^{x-1}.\left(1+5\right)=162\)

\(\Rightarrow3^{x-1}.6=162\)

\(\Rightarrow3^{x-1}=162:6=27=3^3\)

\(\Rightarrow x-1=3\)

\(\Rightarrow x=4\)

Vậy x = 4

Bai 1: Cho tam giac ABC vuong tai A. Tia phan giac cua goc B cat AC  o D. Ke DE vuong goc voi BC .CMR: AB bang BE

Bai 2:  Cho tam giac ABC, D la trung diem cua AB. Duong thang qua D va song2 voi BC cat AC o E, duong thang qua E va song2 voi AB cat BC o F.CMR: 

a, AD bang EF

b, \(\Delta ADE=\Delta EFC\)

c,\(AE=EC\)

Bai 3:* Cho tam giac ABC ,D la trung diem cua AB ,E la trung diem cua AC .Ve diem F : E la trung diem cua DF.CMR:

a,\(DB=CF\)

b,\(\Delta BDC=\Delta FCD\)

c,\(DE//BC,DE=\frac{1}{2}BC\)

HTDT