giúp mình với

a. \(\dfrac{x^2+2x+3}{x^2-x+1}=0\) ⇔x²+2x+3=0 ⇔x²+2x+1+2=0 ⇔(x+1)²+2=0

Vì (x+1)²+2>0 nên phương trình đã cho vô nghiệm.

b) \(\dfrac{x}{x+2}+\dfrac{4}{x-2}=\dfrac{4}{x^2-4}\) ⇔\(\dfrac{x\left(x-2\right)+4\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{4}{\left(x-2\right)\left(x+2\right)}\)

⇔\(x\left(x-2\right)+4\left(x+2\right)=4\) ⇔x²-2x+4x+8-4=0 ⇔x²+2x+4=0                ⇔x²+2x+1+3=0 ⇔(x+1)²+3=0

Vì (x+1)²+3>0 nên phương trình đã cho vô nghiệm.

c: Ta có: \(x^3+3x^2+3x-7=0\)

\(\Leftrightarrow x+1=2\)

hay x=1

b: Ta có: \(x\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

\(\Leftrightarrow x^4+x^3-10x^2+1=x^3-8\)

\(\Leftrightarrow x^4-10x^2+9=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\\x=-3\end{matrix}\right.\)

a: \(\Leftrightarrow14x=56\)

hay x=4

`a,`\(2^x -15= 2^4+1\)

`-> 2^x-15=17`

`-> 2^x=17+15`

`-> 2^x=32`

`-> 2^x=2^5`

`-> x=5`

`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?

`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)

`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)

`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)

`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)

Mà `1/65+1/64-1/63-1/62 \ne 0`

`-> x+66=0`

`-> x=-66`

a: =>2^x=2^4+16=32

=>x=5

b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)

=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)

=>x+66=0

=>x=-66

a)5(x+1)(x-x-2)=0

=>5(x+1).-2=0

=>5(x+1)=0

=>x+1=0

=>x=-1

a)5x.(x+1)-5.(x+1).(x-2)=0

⇒5x(x+1)-(5x-10)(x+1)=0

⇒(x+1)(5x-5x+10)=0

⇒10(x+1)=0

⇒x+1=0⇒x=-1

a) \(5x\left(x+1\right)-5\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow5\left(x+1\right)\left(x-x+2\right)=0\)

\(\Leftrightarrow10\left(x+1\right)=0\)

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

b) \(\left(4x+1\right)\left(x-2\right)-\left(2x-3\right)^2=4\)

\(\Leftrightarrow4x^2-7x-2-4x^2+12x-9=4\)

\(\Leftrightarrow5x=15\Leftrightarrow x=3\)

\(a,2^x=8\\ \Leftrightarrow2^x=2^3\\ \Leftrightarrow x=3\\ b,2^x=\dfrac{1}{4}\\ \Leftrightarrow2^x=2^{-2}\\ \Leftrightarrow x=-2\\ c,2^x=\sqrt{2}\\ \Leftrightarrow2^x=2^{\dfrac{1}{2}}\\ \Leftrightarrow x=\dfrac{1}{2}\)

a) \(\sqrt{x}< 3\)<=> x<9

b)\(\sqrt{4-x}\) ≤ 2 <=> 4 - x ≤ 4 <=> x≥0

c)\(\sqrt{x+2}=\sqrt{4-x}\) <=> x+2=4-x <=>2x=2<=>x=1 

Vậy x=1

d)\(\sqrt{x^2-1}\)=x-1 <=> x\(^2\)-1=x\(^2\)-2x+1 <=> x\(^2\)-\(x^2\)-2x+1+1=0 <=> 2x=2 <=> x=1

Vậy x=1

a) ĐK: x ≥ 0 

⇔ x<9 (TM)

b) ĐK: x ≤ 4 

⇔ 4 - x < 4 

⇔ x > 0

Vậy 0 < x ≤ 4

c) ĐK: -2 ≤ x ≤ 4

Bình phương 2 vế của phương trình, ta có:

x+2=4-x

⇔ 2x = 2

⇔ x=1 (TM)

d) ĐK: x ≥ 1

Bình phương 2 vế của phương trình, ta có:

\(\text{x}^{\text{2}}-11=x^2-2x+1\)

⇔ 2x = 12

⇔ x = 6 (TM)

bn tự kết luận giùm mk nha ^^