`24^2 - 25 + (2x + 5)^2 = 0`

Ta có: `24^2 > 25`

`=> 24^2 - 25 > 0`

Và `(2x + 5)^2 >= 0 ∀x `

`=> 24^2 - 25 + (2x + 5)^2 > 0`

Vậy phương trình đã cho vô nghiệm

\(a,14364:9+20020\)

\(=1596+20020\)

\(=21616\)

\(b,\left(10356\times5-780\right):6\)

\(=\left(51780-780\right):6\)

\(=51000:6\)

\(=8500\)

`c`

\(=51000:6\)

Vậy `9276 : 39 = 237`(dư`33`)

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

`(x^2 - 4sqrt{3}x + 12)/(x - 2sqrt{3}) (x ne 2sqrt{3})`

`= (x^2 - 2x . 2sqrt{3} + (2sqrt{3})^2)/(x - 2sqrt{3}) `

`= ( (x -2 sqrt{3}  )^2)/(x - 2sqrt{3}) `

`= x - 2sqrt{3}`

`(xsqrt{x} - 1)/(x + sqrt{x} + 1) ` với `x > 0; x ne 1`

`= ((sqrt{x})^3 - 1^3)/(x + sqrt{x} + 1)`

`= ((sqrt{x} -1)(x + sqrt{x} + 1))/(x + sqrt{x} + 1)`

`= sqrt{x} -1`

Vậy \(9030:42=215\)

bằng 215

\(\left[\left(-\dfrac{8}{3}\right)^2\right]^{1010}=\left(-\dfrac{8}{3}\right)^{2020}\)

\(\left[\left(\dfrac{7}{3}\right)^{505}\right]^3=\left(\dfrac{7}{3}\right)^{1515}\)

Bài 2: 

\(a,x-\dfrac{3}{10}=\dfrac{7}{15}\cdot\dfrac{3}{5}\\ =>x-\dfrac{3}{10}=\dfrac{7}{25}\\ =>x=\dfrac{7}{25}+\dfrac{3}{10}\\ =>x=\dfrac{29}{50}\\ b.2x+\dfrac{3}{2}=\dfrac{-2}{5}\\ =>2x=\dfrac{-2}{5}-\dfrac{3}{2}\\ =>x=\dfrac{-4}{10}-\dfrac{15}{10}=\dfrac{-19}{10}\\ =>x=\dfrac{-19}{10}:2=-\dfrac{19}{20}\\ c,\left(x-\dfrac{1}{2}\right)^3=-8\\ =>\left(x-\dfrac{1}{2}\right)^3=\left(-2\right)^3\\ =>x-\dfrac{1}{2}=-2\\ =>x=-2+\dfrac{1}{2}\\ =>x=\dfrac{-3}{2}\\ d,\left(\dfrac{7}{5}\right)^x=\dfrac{49}{25}\\ =>\left(\dfrac{7}{5}\right)^x=\left(\dfrac{7}{5}\right)^2\\=>x=2\)

Câu 3:

a: \(\widehat{xOz}+\widehat{yOz}=180^0\)(hai góc kề bù)

=>\(\widehat{yOz}=180^0-\widehat{xOz}=180^0-60^0=120^0\)

b: Ot là phân giác của góc yOz

=>\(\widehat{yOt}=\widehat{zOt}=\dfrac{\widehat{yOz}}{2}=\dfrac{120^0}{2}=60^0\)

Ta có: \(\widehat{xOt}+\widehat{yOt}=180^0\)(hai góc kề bù)

=>\(\widehat{xOt}+60^0=180^0\)

=>\(\widehat{xOt}=120^0\)

c: Ta có: \(\widehat{xOm}=\widehat{yOt}\)(hai góc đối đỉnh)

mà \(\widehat{yOt}=60^0\)

nên \(\widehat{xOm}=60^0\)

Ta có: \(\widehat{xOm}=\widehat{xOz}\left(=60^0\right)\)

=>Ox là phân giác của góc mOz

Câu 1:

b: \(\dfrac{11}{2}\cdot4\dfrac{5}{3}-2\dfrac{5}{3}\cdot\dfrac{11}{2}\)

\(=\dfrac{11}{2}\left(4+\dfrac{5}{3}-2-\dfrac{5}{3}\right)\)

\(=\dfrac{11}{2}\cdot2=11\)

d: \(\left(\dfrac{3}{7}\right)^0\cdot1^{15}+\dfrac{7}{9}:\left(\dfrac{2}{3}\right)^2-\dfrac{4}{5}\)

\(=1\cdot1+\dfrac{7}{9}:\dfrac{4}{9}-\dfrac{4}{5}\)

\(=1-\dfrac{4}{5}+\dfrac{7}{4}=\dfrac{1}{5}+\dfrac{7}{4}=\dfrac{4}{20}+\dfrac{35}{20}=\dfrac{39}{20}\)

Bài 6:

a) Ta có:

\(\sqrt{x+7}\ge0\forall\left(x\ge-7\right)\\ =>2\sqrt{x+7}\ge0\forall\left(x\ge-7\right)\\ =>A=2\sqrt{x+7}-5\ge-5\forall\left(x\ge-7\right)\)

Dấu "=" xảy ra: `x+7=0`

`<=>x=-7` 

b) Ta có:

\(\sqrt{x-8}\ge0\forall\left(x\ge8\right)\\ =>\dfrac{1}{2}\sqrt{x-8}\ge0\forall\left(x\ge8\right)\\ =>A=-12+\dfrac{1}{2}\sqrt{x-8}\ge0\forall\left(x\ge8\right)\)

Dấu "=" xảy ra: `x-8<=>x=8` 

Bài 7:

a: ĐKXĐ: x>=0

\(-\dfrac{1}{4}\sqrt{x}< =0\forall x\) thỏa mãn ĐKXĐ

=>\(B=-\dfrac{1}{4}\sqrt{x}+4< =4\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x=0

b: ĐKXĐ: \(\left[{}\begin{matrix}x>=2\\x< =-2\end{matrix}\right.\)

\(\sqrt{x^2-4}>=0\forall x\) thỏa mãn DKXĐ
=>\(-\dfrac{1}{4}\sqrt{x^2-4}< =0\forall x\) thỏa mãn ĐKXĐ

=>\(B=-\dfrac{1}{4}\sqrt{x^2-4}+1< =1\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(x^2-4=0\)

=>\(x^2=4\)

=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)