Lời giải:
a. $x^3=4^3\Rightarrow x=4$

b. $x^2=49=7^2=(-7)^2$

$\Rightarrow x=7$ hoặc $x=-7$

c. $x^3+1=28$

$x^3=28-1=27=3^3$

$\Rightarrow x=3$

d. $2^x=16=2^4$

$\Rightarrow x=4$

e. $2^4.2^x=2^6$

$\Rightarrow 2^{4+x}=2^6$

$\Rightarrow 4+x=6$

$\Rightarrow x=2$

g.

$5^x=25.5^3=5^2.5^3=5^5$

$\Rightarrow x=5$

Lần sau bạn lưu ý viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để đề được rõ ràng hơn nhé.

\(\left(x^2+5\right)\left(x^2-49\right)=0\)

\(\left(x^2-49\right)=0\)  vì   \(x^2+5>0\)

\(\Rightarrow x^2=49\)

\(\Rightarrow x=\pm7\)

vậy \(x=\pm7\)

\(\left(x^2+5\right)\left(x^2-7\right)< 0\)

\(\left(x^2-7\right)< 0\)

\(\left(x-\sqrt{7}\right)\left(x+\sqrt{7}\right)< 0\)

\(\hept{\begin{cases}x-\sqrt{7}>0\\x+\sqrt{7}< 0\end{cases}}\)  hoặc \(\hept{\begin{cases}x-\sqrt{7}< 0\\x+\sqrt{7}>0\end{cases}}\)

\(\hept{\begin{cases}x>\sqrt{7}\\x< -\sqrt{7}\end{cases}}\)  hoặc \(\hept{\begin{cases}x< \sqrt{7}\\x>-\sqrt{7}\end{cases}}\)

đến đây tự làm tiếp

\(\left(x^2+5\right)\times\left(x^2-49\right)=0\)

Vậy \(\orbr{\begin{cases}x^2+5=0\\x^2-49=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=-5\\x^2=49\end{cases}}\)

Mà \(x^2\ge0\Rightarrow x^2=-5\)(loại)

Vậy \(x^2=49\)

Nên \(\orbr{\begin{cases}x=7\\x=-7\end{cases}}\)

\(\left(x^2+5\right)\times\left(x^2-7\right)< 0\)

\(\Leftrightarrow\hept{\begin{cases}x^2+5< 0\\x^2-7>0\end{cases}}\)hoặc \(\hept{\begin{cases}x^2+5>0\\x^2-7< 0\end{cases}}\)

TH1: \(\hept{\begin{cases}x^2+5< 0\\x^2-7>0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2< -5\\x^2>7\end{cases}}\)

Mà \(x^2\ge0\Rightarrow x^2< -5\)(loại)

Vậy \(x^2>7\)

TH2: \(\hept{\begin{cases}x^2+5>0\\x^2-7< 0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x^2>-5\\x^2< 7\end{cases}}\)

.....

tới đây thì tự làm nhé, mình bí khúc này

a) x²-24x​-25=x²+x-25x-25

=x(x+1)-25(x+1)

=(x-25)(x+1)

b) x²-y²+14x​+49=(^​x²​+14x+49)-y²

=(x+7)²​-y²​=(x-y+7)(x+y+7)

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

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\Leftrightarrow x=1;x=-1;x=-3\)

b, \(15x-5+6x^2-2x=0\Leftrightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)

\(\Leftrightarrow\left(2x+5\right)\left(3x-1\right)=0\Leftrightarrow x=-\frac{5}{2};x=\frac{1}{3}\)

c, \(5x-2-25x^2+10x=0\)

\(\Leftrightarrow\left(5x-2\right)-5x\left(5x-2\right)=0\Leftrightarrow\left(1-5x\right)\left(5x-2\right)=0\Leftrightarrow x=\frac{2}{5};x=\frac{1}{5}\)

= 1/5 nha

a) \(\left(6x-5y\right)^2=36x^2-60xy+25y^2\)

b) \(\left(4x-1\right)^2=16x^2-8x+1\)

c) \(\left(x+2\right)^2=x^2+4x+4\)

d) \(x^2-64=\left(x-8\right)\left(x+8\right)\)

e) \(4x^2-64=\left(2x-8\right)\left(2x+8\right)\)

f) \(25x^2-4=\left(5x-2\right)\left(5x+2\right)\)

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

h) \(\left(x-3\right)^3=x^3-9x^2+27x-27\)

k) \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)

l) \(x^3-125=\left(x-5\right)\left(x^2+5x+25\right)\)

y) \(27y^3-1=\left(3y-1\right)\left(9y^2+3y+1\right)\)

^{ai Giúp mình với}

ko có đề bài hả bn

1. \(4x^2-49=0\)

\(\Leftrightarrow\left(2x+7\right)\left(2x-7\right)=0\)

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

Vậy: \(x=-\dfrac{7}{2}\) hoặc \(x=\dfrac{7}{2}\)

===========

2. \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x=6\)

Vậy: \(x=6\)

===========

3. \(10\left(x-5\right)-8x\left(5-x\right)=0\)

\(\Leftrightarrow10\left(x-5\right)+8x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(10+8x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\Leftrightarrow x=5\\10+8x=0\Leftrightarrow x=-\dfrac{5}{4}\end{matrix}\right.\)

Vậy: \(x=5\) hoặc \(x=-\dfrac{5}{4}\)

1: Ta có: \(4x^2-49=0\)

\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

2: Ta có: \(x^2+36=12x\)

\(\Leftrightarrow x^2-12x+36=0\)

\(\Leftrightarrow\left(x-6\right)^2=0\)

\(\Leftrightarrow x-6=0\)

hay x=6

a, 6

b,1

c,1

a;2 mu x =2 mu 6 .Vậy x=6