\(a,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+6\sqrt{x-1}=14\\ \Leftrightarrow7\sqrt{x-1}=14\\ \Leftrightarrow\sqrt{x-1}=2\Leftrightarrow x-1=4\\ \Leftrightarrow x=5\left(tm\right)\\ b,ĐK:-2\le x\le2\\ PT\Leftrightarrow\sqrt{2-x}\left(1-\sqrt{2+x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2-x=0\\2+x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

a) ĐKXĐ: \(x\ge1\)

\(pt\Leftrightarrow\sqrt{x-1}+6\sqrt{x-1}=14\)

\(\Leftrightarrow7\sqrt{x-1}=14\Leftrightarrow\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\)

b) ĐKXĐ: \(-2\le x\le2\)

\(pt\Leftrightarrow\sqrt{2-x}-\sqrt{\left(2-x\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\sqrt{2-x}\left(1-\sqrt{x+2}\right)=0\)

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

dài quá à :(

bạn lm 5 câu cuối cũng được rùi

bấm chữ 0 đúng sẽ ra câu trả lời .

bấm vào chữ 0 đúng sẽ ra đáp án 

a)4x+4-3x+1=14

x+5=14

x=11

b)trường hợp 1  x²-9=0 

                        x²=9

->x=3;-3

-trường hợp 2: x+2=0

x=-2

c)-th1:x²+9=0

x²=-9

->x rỗng

d)xy+2x-y-2=0

(xy-y)+(2x-2)=0

y(x-1)+2(x-1)=0

(y+2)(x-1)=0

th1: y+2=0

y=-2

th2:x-1=0

x=1

(th1: trường hợp 1)

1) \(2x\left(x-3\right)+5x-15=0\)

\(2x\left(x-3\right)+5\left(x-3\right)=0\)

\(\left(x-3\right)\left(2x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-5}{2}\end{matrix}\right.\)

2) \(x\left(2x-7\right)-4x+14=0\)

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

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

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

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

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

\(x-6=0\)

\(x=6\)

4) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)

\(\left(x^3+3^3\right)-x\left(x^2-1\right)-27=0\)

\(x^3+27-x^3+x-27=0\)

\(x=0\)

a. Ta có: x²-11=0

⇌ x²=11

⇌\(\left[{}\begin{matrix}x=\sqrt{11}\\x=-\sqrt{11}\end{matrix}\right.\)

b.Ta có: x²-2\(\sqrt{13}\)x+\(\sqrt{13}\)=0

⇌(x-\(\sqrt{13}\))²=0

⇌ x-\(\sqrt{13}\)=0

⇌ x=\(\sqrt{13}\)

c. Ta có : x²-9x+14=0

⇌ (x-7)(x-2)=0

⇌\(\left[{}\begin{matrix}x-7=0\\z-2=0\end{matrix}\right.\)⇌\(\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)

d.Ta có \(\sqrt{x}\)-6=13

⇌\(\sqrt{x}\)=19

⇌x = 361

e.Ta có: \(\sqrt{x}\)+9=3

Vì \(\sqrt{x}\)≥0∀x⇒\(\sqrt{x}\)+9≥9

⇒ ptvn

f.Ta có:\(\sqrt{x^2}\)-2x+4=x-1

⇌ |x|-3x-5=0(*)

TH1: x≥0

⇒ pt(*) ⇌ x-3x+5=0⇌-2x-5=0⇒x=\(\dfrac{5}{2}\)(t/m)

TH2: x<0

⇒ pt(*) ⇌ -x-3x+5=0⇌-4x+5=0⇒x=\(\dfrac{5}{4}\)(l)

Vậy x=\(\dfrac{5}{2}\)là nghiệm của phương trình