\(3^{x+2}+4\cdot3^{x+1}=7\cdot3^6\\ 3^x\cdot3^2+4\cdot3^x\cdot3=5103\\ 3^x\left(9+12\right)=5103\\ 3^x\cdot21=5103\\ 3^x=243\\ 3^x=3^5\\ x=5\)

haidangbb ông cố ơi còn câu b ạ

A,X=963-654

x=309

x³-3x=0

=>x=0

tik nha ban

a,x+654=963

x=963-654

x=309

b,x =0

\(4.3^{x+2}-3^{x-1}=963\)

\(\Rightarrow4.3^{x+2}-3^{x+2-3}=963\)

\(\Rightarrow4.3^{x+2}-\dfrac{1}{3^3}3^{x+2}=963\)

\(\Rightarrow4.3^{x+2}-\dfrac{1}{27}3^{x+2}=963\left(1\right)\)

Đặt \(t=3^{x+2}>0\)

\(\left(1\right)\Rightarrow4t-\dfrac{1}{27}t=963\)

\(\Rightarrow t\left(4-\dfrac{1}{27}\right)=963\)

\(\Rightarrow\dfrac{107}{27}.t=9.107\)

\(\Rightarrow t=3^2\Rightarrow t=3^2:\dfrac{1}{3^3}=3^2.\dfrac{3^3}{1}=3^5\)

\(\Rightarrow3^{x+2}=3^5\Rightarrow x+2=5\Rightarrow x=3\)

4\(\cdot\)3^x+2-3^x-1=963

=>4\(\cdot\)3^xx3²-3^x:3=963

=>3^x\(\cdot\)36-3^x\(\cdot\)\(\dfrac{1}{3}\)=963

=>3^x\(\cdot\)(36-\(\dfrac{1}{3}\))=963

=>3^x\(\cdot\)\(\dfrac{107}{3}\)=963

=>3^x=963:\(\dfrac{107}{3}\)=27=3³

=>x=3

Chọn A.

a. 113 + 142 + x = 999 – 103

255 + x = 896

x = 896 – 255

x = 641

b. x – 124 = 400 + 56

x – 124 = 456

x = 456 + 124

x = 580

c. 963 – x = 869 – 28 : 4

963 – x = 869 – 7

963 – x = 862

x = 963 – 862

x = 101

Ta có: \(4.3^{x+2}-2.3^{x+1}=810\)

\(4.3^{x+1}.3-2.3^{x+1}=810\)

\(2.3^{x+1}\left(2.3-1\right)=810\)

\(2.3^{x+1}.5=810\)

\(3^{x+1}=81=3^4\)

x+1=4

x=3

\(\left(3x+1\right)^2=25\)

\(\Rightarrow\left(3x+1\right)^2=5^2=\left(-5\right)^2\)

\(\Rightarrow\orbr{\begin{cases}3x+1=5\\3x+1=-5\end{cases}\Rightarrow\orbr{\begin{cases}3x=5-1=4\\3x=-5-1=-6\end{cases}}}\Rightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-2\end{cases}}\)

\(\left[x-\frac{1}{2}\right]+\frac{1}{2}=\frac{5}{8}\)

\(\Rightarrow x-0=\frac{5}{8}\)

\(x=\frac{5}{8}\)

\(\left[x+\frac{3}{4}\right]-\frac{1}{3}=0\)

\(x+\frac{3}{4}=0+\frac{1}{3}=\frac{1}{3}\)

\(x=\frac{1}{3}-\frac{3}{4}\)

\(x=\frac{-5}{12}\)

\(\frac{1}{9}.3^4.3x=3^7\)

\(\Rightarrow\frac{1}{9}.3^5x=3^7\)

\(x=3^7:\frac{1}{9}:3^5\)

\(x=3^2.9\)

\(x=81\)

\(\frac{x}{5}=\frac{4}{21}\)

\(\Rightarrow21x=4.5\)

\(\Rightarrow x=\frac{4.5}{21}\)

\(x=\frac{20}{21}\)

Đặt \(3^x=a\)  \(\left(a>0\right)\)

Phương trình \(\Leftrightarrow a^2-4a+m-2=0\) (*)

Yêu cầu bài toán \(\Leftrightarrow\) Phương trình (*) có 2 nghiệm dương phân biệt

\(\Leftrightarrow\left\{{}\begin{matrix}\Delta'>0\\a_1+a_2>0\\a_1\cdot a_2>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}4-\left(m-2\right)>0\\4>0\left(t/m\right)\\m-2>0\end{matrix}\right.\) \(\Leftrightarrow\) ...