\(a,2^{x+1}=64\\ \Rightarrow a,2^{x+1}=2^6\\ \Rightarrow x+1=6\\ \Rightarrow x=5\)

\(b,x=18\)

\(c,\left(4x-9\right)-\left(x+111\right)=0\\ \Rightarrow4x-9-x-111=0\\ \Rightarrow3x-120=0\\ \Rightarrow3x=120\\ \Rightarrow x=40\)

a
,
2
x
+
1
=
64
⇒
a
,
2
x
+
1
=
2
6
⇒
x
+
1
=
6
⇒
x
=
5

b
,
x
=
18

c
,
(
4
x
−
9
)
−
(
x
+
111
)
=
0
⇒
4
x
−
9
−
x
−
111
=
0
⇒
3
x
−
120
=
0
⇒
3
x
=
120
⇒
x
=
40

\(\left(x^2-9\right)^2-9\left(x-3\right)^2=0\)

\(< =>\left(x^2-9\right)^2-\left[3\left(x-3\right)\right]^2=0\)

\(< =>\left(x^2-9\right)^2-\left(3x-9\right)^2=0\)

\(< =>\left(x^2-9+3x-9\right)\left(x^2-9-3x+9\right)=0\)

\(< =>\left(x^2+3x-18\right)\left(x^2-3x\right)=0\)

\(=>\left[{}\begin{matrix}x^2+3x-18=0\\x^2-3x=0\end{matrix}\right.< =>\left[{}\begin{matrix}\left(x+6\right)\left(x-3\right)=0\\x\left(x-3\right)=0\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x=-6\\x=3\\x=0\end{matrix}\right.\)

\(a,\Leftrightarrow x^2-2x-x^2+5x=6\\ \Leftrightarrow3x=6\\ \Leftrightarrow x=2\)

\(b,\Leftrightarrow x^2-6x+9-x+9=0\\ \Leftrightarrow x^2-7x+18=0\\ \Leftrightarrow\left(x^2-7x+\dfrac{49}{4}\right)+\dfrac{23}{4}=0\\ \Leftrightarrow\left(x-\dfrac{7}{2}\right)^2+\dfrac{23}{4}=0\left(vôlí\right)\)

\(a,1-\left(\dfrac{\dfrac{5}{3}}{8}+x-\dfrac{\dfrac{7}{5}}{24}\right)-\dfrac{\dfrac{16}{2}}{3}=0\\ \Leftrightarrow1-\left(\dfrac{5}{24}+x-\dfrac{7}{120}\right)=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{3}{20}+x=1-\dfrac{8}{3}=-\dfrac{5}{3}\\ \Leftrightarrow x=-\dfrac{5}{3}-\dfrac{3}{20}=-\dfrac{109}{60}\)

Ta có: \(1-\left(\dfrac{5}{\dfrac{3}{8}}+\dfrac{x-7}{\dfrac{5}{24}}\right)-\dfrac{16}{\dfrac{2}{3}}=0\)

\(\Leftrightarrow\dfrac{40}{3}+\dfrac{24\left(x-7\right)}{5}=23\)

\(\Leftrightarrow\left(x-7\right)\cdot\dfrac{24}{5}=\dfrac{29}{3}\)

\(\Leftrightarrow x-7=\dfrac{145}{72}\)

hay \(x=\dfrac{649}{72}\)

a. \(x^2-25-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5\right)-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b. \(\left(3x+1\right)^2=\left(2x-5\right)\\ \Leftrightarrow9x^2+6x+1=2x-5\\ \Leftrightarrow9x^2+6x-2x=-5-1\\ \Leftrightarrow9x^2+4x=-6\\ \Leftrightarrow x\left(9x+4\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\9x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-\dfrac{10}{9}\end{matrix}\right.\)

c. \(2x^2-7x+6=0\\ \Leftrightarrow2x^2-7x=-6\\ \Leftrightarrow x\left(2x-7\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{1}{2}\end{matrix}\right.\)

a, \(\left(x-5\right)\left(x+5\right)-3\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=-2;x=5\)

b, bạn ktra lại đề, thường thường ngta hay cho 2 vế cùng bình phương 

c, \(2x^2-7x+6=0\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\Leftrightarrow x=\dfrac{3}{2};x=2\)

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

\(\left[{}\begin{matrix}3x+1=2x-5\\3x+1=5-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\5x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{4}{5}\end{matrix}\right.\)

=> x< 2 và x>-9

Đây là lớp 6 nhé ko phải lớp 8 đâu. Tại mình ghi nhầm

Vì \(x^2+1>0\) nên \(x^2-4=0\)

\(\Leftrightarrow x^2=4\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

x=-2 và 2