Question

Bai1 tìm x. giải từng bước giúp mình na...

a \(\sqrt{9x}=15\)

b \(\sqrt{4x^2}=8\)

c \(\sqrt{9\left(2x-3\right)^2}=6\)

d \(\sqrt{4\left(x+1\right)}=\sqrt{8}\)              

e \(\sqrt{x^2-4}-\sqrt{x-2}=0\)

f \(\sqrt{x+1}+\sqrt{9x+9}=4\)

g \(\sqrt{x-2}-\sqrt{x\left(x-2\right)}=0\)

h \(\sqrt{x^2-9}-\sqrt{x+3}=0\)

k \(\sqrt{x+2}.\sqrt{2x-4}+\sqrt{x-2}.\sqrt{2x-4}=2\sqrt{2}\)

Answer 1

a. \(9x=225\Rightarrow x=25\)

b. \(2x=8\Rightarrow x=4\)

c. \(3\left(2x-3\right)=6\Rightarrow6x=15\Rightarrow x=\frac{15}{6}\)

d. \(4\left(x+1\right)=8\Rightarrow4x=4\Rightarrow x=1\)

e. \(\sqrt{x+2}.\sqrt{x-2}-\sqrt{x-2}=0\Rightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)

=> \(\sqrt{x-2}=0\Rightarrow x-2=0\Rightarrow x=2\)

hoặc \(\sqrt{x+2}-1=0\Rightarrow\sqrt{x+2}=1\Rightarrow x+2=1\Rightarrow x=-1\)

f. \(\sqrt{x+1}+3\sqrt{x+1}=4\Rightarrow4\sqrt{x+1}=4\Rightarrow\sqrt{x+1}=1\Rightarrow x+1=1\Rightarrow x=0\)

g. \(\sqrt{x-2}\left(1-\sqrt{x}\right)=0\)

=> \(\sqrt{x-2}=0\Rightarrow x-2=0\Rightarrow x=2\)

hoặc \(1-\sqrt{x}=0\Rightarrow\sqrt{x}=1\Rightarrow x=1\)

h. \(\sqrt{x+3}.\sqrt{x-3}-\sqrt{x+3}=0\Rightarrow\sqrt{x+3}\left(\sqrt{x-3}-1\right)=0\)

=> \(\sqrt{x+3}=0\Rightarrow x=-3\)

hoặc \(\sqrt{x-3}-1=0\Rightarrow\sqrt{x-3}=1\Rightarrow x=4\)

 

 

 

 

 

 

 

Answer 2

√x−2(1−√x)=0

=> √x−2=0⇒x−2=0⇒x=2

hoặc 1−√x=0⇒√x=1⇒x=1

h. √x+3.√x−3−√x+3=0⇒√x+3(√x−3−1)=0

=> √x+3=0⇒x=−3

hoặc 

√x−2(1−√x)=0

=> √x−2=0⇒x−2=0⇒x=2

hoặc 1−√x=0⇒√x=1⇒x=1

h. √x+3.√x−3−√x+3=0⇒√x+3(√x−3−1)=0

=> √x+3=0⇒x=−3

hoặc 

Answer 3

khong biet minh cung khong