a)

Vì 5^2 =25

=> \(\sqrt{25}=5\)

Vì 7^2=49

=>\(\sqrt{49}=7\)

Vì 1^2=1

=>\(\sqrt{1}=1\)

Vì \(\left(\frac{2}{3}\right)^2=\frac{4}{9}\)

=> \(\sqrt{\frac{4}{9}}=\frac{2}{3}\)

a) Vì \(5^2=25\Rightarrow\sqrt{25}=5\)

b) Vì \(7^2=49\Rightarrow\sqrt{49}=7\)

c) Vì \(1^2=1\Rightarrow\sqrt{1}=1\)

d) Vì \(\left(\frac{2}{3}\right)^2=\frac{4}{9}\Rightarrow\sqrt{\frac{4}{9}}=\frac{2}{3}\)

a)Vì 5²=25 nên \(\sqrt{25}=5\)

b)Vì 7²=49 nên \(\sqrt{49}=7\)

c)Vì 1²=1 nên \(\sqrt{1}=1\)

d)Vì \(\left(\frac{2}{3}\right)^2=\frac{4}{9}\) nên \(\sqrt{\frac{4}{9}}=\frac{\sqrt{4}}{\sqrt{9}}=\frac{2}{3}\)

a) Vì 52 = 25 nên √25 = 5

b) Vì 72 = 49 nên √49 = 7

a) Vì 5² = 25 nên √25 = 5

b) Vì 7²= 49 nên √49 = 7

c) Vì 1² = 1 nên √1 = 1

d) Vì = nên

a) Vì 5²=25 nên \(\sqrt{25}=5\).

b) Vì 7²=49 nên \(\sqrt{49}=7\).

c) Vì 1ⁿ=1 nên \(\sqrt{1}=1\). (\(\forall n\in N\))

d) Vì \(\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) nên \(\sqrt{\dfrac{4}{9}}=\dfrac{\sqrt{4}}{\sqrt{9}}=\dfrac{2}{3}\).

a) Vì 5² = 25 nên √25 = 5

b) Vì 7²= 49 nên √49 = 7

c) Vì 1² = 1 nên √1 = 1

d) Vì (23)2=49(23)2=49 = nên √49=23

ĐKXĐ: \(\left\{{}\begin{matrix}2x+5>=0\\4-2x>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x>=-5\\2x< =4\end{matrix}\right.\Leftrightarrow-\dfrac{5}{2}< =x< =2\)

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

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

=>\(\left(x-2\right)\left(x+2\right)+\dfrac{2x+5-9}{\sqrt{2x+5}+3}+\sqrt{4-2x}=4x-8\)

=>\(\left(x-2\right)\left[\left(x+2\right)+\dfrac{2}{\sqrt{2x+5}+3}-4\right]+\sqrt{4-2x}=0\)

=>\(-\left(2-x\right)\left[\left(x-2\right)+\dfrac{2}{\sqrt{2x+5}+3}\right]+\sqrt{2\left(2-x\right)}=0\)

=>\(\sqrt{2-x}\left[-\sqrt{2-x}\left(x-2+\dfrac{2}{\sqrt{2x+5}+3}\right)+\sqrt{2}\right]=0\)

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

=>x=2(nhận)

giúp mình nha mình cần gấp cảm ơn các b nhiều

Vì 5² = 25 nên\(\sqrt{25}=5\);vì 7² = 49 nên\(\sqrt{49}=7\)

a: \(F\left(-2\right)=\dfrac{3}{2}\cdot\left(-2\right)^2=\dfrac{3}{2}\cdot4=6\)

F(3)=3/2*3^2=27/2

\(F\left(\sqrt{5}\right)=\dfrac{3}{2}\cdot\left(\sqrt{5}\right)^2=\dfrac{3}{2}\cdot5=\dfrac{15}{2}\)

\(F\left(-\dfrac{\sqrt{2}}{3}\right)=\dfrac{3}{2}\cdot\dfrac{2}{9}=\dfrac{3}{9}=\dfrac{1}{3}\)

b: \(F\left(-2\right)=\dfrac{3}{2}\cdot\left(-2\right)^2=\dfrac{3}{2}\cdot4=6\)

=>A thuộc (P)

\(F\left(-\sqrt{2}\right)=\dfrac{3}{2}\cdot\left(-\sqrt{2}\right)^2=\dfrac{3}{2}\cdot2=3\)

=>B thuộc (P)

\(F\left(-4\right)=\dfrac{3}{2}\cdot\left(-4\right)^2=\dfrac{3}{2}\cdot16=\dfrac{48}{2}=24\)

=>C ko thuộc (P)

F(1/căn 2)=3/2*1/2=3/4

=>D thuộc (P)