\(c.\:\left(3x+4\right)^2-\left(3x+1\right)\left(3x-1\right)\\ =9x^2+24x+16-9x^2+1\\ 40x=-1\\ x=-\dfrac{1}{40}\)

\(d.\:\left(3x-1\right)^2-\left(3x-2\right)^2=0\\ \left(3x-1+3x-2\right)\left(3x-1-3x+2\right)=0\\ \left(6x-3\right)=0\\ x=\dfrac{1}{2}\)

\(g.\:\left(2x+1\right)^2-\left(x-1\right)^2=0\\ \left(2x+1+x-1\right)\left(2x+1-x+1\right)=0\\ 3x\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

c,\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)

\(\Rightarrow9x^2+24x+16-\left(9x^2-1\right)=49\)

\(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=49-1-16\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

d, \(\left(3x-1\right)^2-\left(3x-2\right)^2=0\)

\(\Rightarrow\left(3x-1-3x+2\right).\left(3x-1+3x-2\right)=0\)

\(\Rightarrow6x-3=0\Rightarrow6x=3\Rightarrow x=\dfrac{1}{2}\)

e, \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)

\(\Rightarrow\left(2x+1-x+1\right)\left(2x+1+x-1\right)=0\)

\(\Rightarrow\left(x+2\right).3x=0\Rightarrow x.\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Chúc bạn học tốt!!!

a: 5-3x=6x+7

=>-3x-6x=7-5

=>-9x=2

=>\(x=-\dfrac{2}{9}\)

b: \(\dfrac{3x-2}{6}-5=3-\dfrac{2\left(x+7\right)}{4}\)

=>\(\dfrac{3x-2}{6}+\dfrac{x+7}{2}=8\)

=>\(\dfrac{3x-2+3\left(x+7\right)}{6}=8\)

=>3x-2+3x+14=48

=>6x+12=48

=>6x=36

=>\(x=\dfrac{36}{6}=6\)

c: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

=>\(\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)

=>(x-1)(5x+3-3x+8)=0

=>(x-1)(2x+11)=0

=>\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)

d: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)

=>\(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)

=>\(\left(x-4\right)\left(3x+2\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)

xem lại câu  1 đi

a.

\(2^x=2^{3x-1}\Leftrightarrow x=3x-1\)

\(\Rightarrow x=\dfrac{1}{2}\)

b.

\(7^{x-5}=49\Leftrightarrow x-5=log_749=2\)

\(\Rightarrow x=7\)

c.

\(3^{5x-3}=1\Rightarrow5x-3=log_31=0\)

\(\Rightarrow x=\dfrac{3}{5}\)

d.

\(\left(\dfrac{1}{7}\right)^{5x}=7^{x+6}\Leftrightarrow7^{-5x}=7^{x+6}\)

\(\Leftrightarrow-5x=x+6\)

\(\Rightarrow x=-1\)

b)(x-2)3-(x-3)(x2+3x+9)+6(x+1)2=49

(=) x³- 6x² +12 x -8 - ( x3 - 27 ) + 6( x² + 2x +1)

(=) x³ - 6x² +12x -8 - x³ +27 + 6x² +12x +6

(=) 24x + 25 = 49

(=) 24x = 49 - 25 = 24

(=) x = 24/24 =1

a, \(A=x^2+2\cdot\frac{1}{2}x+\frac{1}{4}-\frac{9}{4}=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\)

=> \(A\ge-\frac{9}{4}\) dấu = xảy ra khi : \(x=\frac{-1}{2}\)

b, \(B=x^2-2.\frac{1}{2}.x+\frac{1}{4}-\frac{1}{4}=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\)

=> \(B\ge-\frac{1}{4}\) dấu = <=> \(x=\frac{1}{2}\)

c, \(C=\frac{1}{4}.x^2-2.\frac{1}{2}x+1+6=\left(\frac{1}{2}x^{ }-1\right)^2+6\)

=> \(C\ge6\) dấu = <=> \(x=2\)

b) (x+1)^3-x(x-2)^2+x-1=0

 ⇔x^3+3x^2+3x+1-(x^3-4x^2+4x)=0

⇔ x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0

⇔7x^2-2=0

⇔7x^2=2

⇔7x^2=-2⇔x=-3

⇔7x^2=2⇔x=-căn 5