Question

Giải phương trình:

1)\(\sqrt{x}+\sqrt{x+1}=1\)

2) \(\sqrt{x+2\sqrt{2x-5}-2}+\sqrt{x-3\sqrt{2x-5}+2}=2\sqrt{x}\)

3) \(\sqrt{x\left(3x+1\right)}-\sqrt{x\left(x-1\right)}=2\sqrt{x^2}\)

4) \(4\sqrt{x+1}=x^2-5x+14\)

5) \(\sqrt{x^2-7x+3}-\sqrt{x^2-2}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}\)

 

 

Answer 1

can x = 1 - can x+1

bphuong

x^2=2+2 can x+1

Answer 2

1)\(\left(DKXD:x\ge0\right)\)

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

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

\(\Leftrightarrow x\left(x+1\right)=1-2x+x^2\left(0\le x\le1\right)\)

\(\Leftrightarrow x^2+x=1-2x+x^2\)

\(\Leftrightarrow3x-1=0\)

\(\Leftrightarrow x=\frac{1}{3}\)

Vậy pt có nghiệm \(x=\frac{1}{3}\)

Answer 3

1/3 nha

Answer 4

2x+2√​x(x+1)​​​+1=1

2x+2√​x(x+1)​​​=0

2√​x(x+1)​​​=−2x

4x(x+1)=4x​2​​

x(x+1)=x​2​​

x​2​​+x=x​2

x=0

b) tự làm sai thì thôi vì mới học lớp 7 ok 

Answer 5

câu 4 làm sao vậy giúp tui với :v

Answer 6

4.\(4\sqrt{x+1}=x^2-5x+14\)

ĐK \(x\ge-1\)

pT <-> \(\left(x^2-6x+9\right)+\left(x+1-4\sqrt{x+1}+4\right)=0\)

<=> \(\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)

Do \(VT\ge0\)

=> \(\hept{\begin{cases}x=3\\\sqrt{x+1}=2\end{cases}\Rightarrow}x=3\left(tmĐK\right)\)

Vậy x=3