\(\sqrt{7-2\sqrt{10}}=\sqrt{5-2\sqrt{10}+2}\\ =\sqrt{\sqrt{5}^2-2.\sqrt{5}.\sqrt{2}+\sqrt{2}^2}\\ =\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\left|\sqrt{5}-\sqrt{2}\right|\\ =\sqrt{5}-\sqrt{2}\)

\(\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{2}.\sqrt{5}+\left(\sqrt{2}\right)^2}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\left(vì\sqrt{5}-\sqrt{2}>0\right)\)

Lời giải:

$\sqrt{11-2\sqrt{10}}=\sqrt{10-2\sqrt{10}+1}=\sqrt{(\sqrt{10}-1)^2}$

$=|\sqrt{10}-1|=\sqrt{10}-1$

P/s: Lần sau bạn lưu ý đăng đầy đủ yêu cầu đề bài.

\(\sqrt{11-2\sqrt{10}}=\sqrt{10-2\sqrt{10}+1}\\ =\sqrt{\sqrt{10}^2-2.\sqrt{10}.1+1^2}\\ =\sqrt{\left(\sqrt{10}-1\right)^2}\\ =\left|\sqrt{10}-1\right|=\sqrt{10}-1\)

`a)`\(\sqrt{3x+1}-\sqrt{x-1}=2\)

\(ĐK:x\ge1\)

\(\Leftrightarrow3x+1+x-1-2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)

\(\Leftrightarrow4x-2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)

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

\(\Leftrightarrow\left(3x+1\right)\left(x-1\right)=4x^2-8x+4\)

\(\Leftrightarrow3x^2-3x+x-1=4x^2-8x+4\)

\(\Leftrightarrow x^2-6x+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\) `(tm)`

Vậy \(S=\left\{5;1\right\}\)

`b)`\(\dfrac{\sqrt{x+27}+\sqrt{27-x}}{\sqrt{27+x}-\sqrt{27-x}}=\dfrac{27}{x}\)

\(ĐK:-27\le x\le27;x\ne0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+27}+\sqrt{27-x}\right)\left(\sqrt{x+27}-\sqrt{27-x}\right)}{\left(\sqrt{27+x}-\sqrt{27-x}\right)\left(\sqrt{27+x}-\sqrt{27-x}\right)}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+27}\right)^2-\left(\sqrt{27-x}\right)^2}{\left(\sqrt{27+x}-\sqrt{27-x}\right)^2}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{2x}{54-2\sqrt{\left(27+x\right)\left(27-x\right)}}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{x}{27-\sqrt{27^2-x^2}}=\dfrac{27}{x}\)
\(\Leftrightarrow x^2=27^2-\sqrt{27^2-x^2}\)

\(\Leftrightarrow27^2-x^2-\sqrt{27^2-x^2}=0\)

\(\Leftrightarrow\sqrt{27^2-x^2}\left(\sqrt{27^2-x^2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}27^2-x^2=0\\\sqrt{27^2-x^2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm27\\x=\pm\sqrt{27^2-1}\end{matrix}\right.\)

Thế vào pt ta được \(x=\pm27\) là thỏa mãn

Vậy \(S=\left\{\pm27\right\}\)

a) ĐKXĐ: \(x\ge1\)

Ta có: \(\sqrt{3x+1}-\sqrt{x-1}=2\)

\(\Leftrightarrow\sqrt{3x+1}-4+2-\sqrt{x-1}=0\)

\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{4-x+1}{2+\sqrt{x-1}}=0\)

\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{5-x}{2+\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}-\dfrac{1}{2+\sqrt{x-1}}\right)=0\)

+ Nếu \(x-5=0\Leftrightarrow x=5\left(tmđkxđ\right)\)

+ Nếu \(\dfrac{3}{\sqrt{3x+1}+4}-\dfrac{1}{2+\sqrt{x-1}}=0\Leftrightarrow\dfrac{3}{\sqrt{3x+1}+4}=\dfrac{1}{2+\sqrt{x-1}}\)

\(\Leftrightarrow3\sqrt{x-1}+2=\sqrt{3x+1}\Rightarrow6x-6+12\sqrt{x-1}=0\)

\(\sqrt{x-1}\left(6\sqrt{x-1}+12\right)=0\Leftrightarrow\sqrt{x-1}=0\)

\(\Leftrightarrow x=1\left(tmđkxđ\right)\)

Vậy ....

a)1000-385-296+78

= 615 - 296 + 78 

= 319 + 78 

= 397 

b)90:5x3

= 18 x 3 

=  54

c)600:3:2x5

= 200:2 x 5 

= 100 x 5 

= 500 

a) 1000 - 385 - 296 + 78

=  615 - 296 + 78

= 319 + 78

= 397

b) 90 : 5 x 3

= 18 x 3

= 54

c) 600 : 3 : 2 x 5

= 200 : 2 x 5

= 100 x 5

= 500

Lời giải:

$A=1+3+3^2+3^3+...+3^{2022}$

$=1+(3+3^2)+(3^3+3^4)+...+(3^{2021}+3^{2022})$

$=1+3(1+3)+3^3(1+3)+....+3^{2021}(1+3)$

$=1+4(3+3^3+...+3^{2021})$

$=1+4[3+(3^3+3^5)+(3^7+3^9)+....+(3^{2019}+3^{2021})]$

$=13+4[3^3(1+3^2)+3^7(1+3^2)+...+3^{2019}(1+3^2)]$
$=13+40(3^3+3^7+...+3^{2019})$
$=13+40[3^3+(3^7+3^{11})+(3^{15}+3^{19})+...+(3^{2015}+3^{2019})]$
$=1093+40[3^7(1+3^4)+3^{15}(1+3^4)+....+3^{2015}(1+3^4)]$

$=1093+40.82(3^7+3^{15}+...+3^{2015})$
$=1093+16.5.41(3^7+...+3^{2015})$

$=5+16.68+16.5.41(3^7+...+3^{2015})$

Vậy biểu thức chia $16$ dư $5$

Số phần gạo cửa hàng ấy đã bán được cả buổi sáng và buổi chiều là:

           \(\dfrac{1}{7}+\dfrac{2}{7}=\dfrac{3}{7}\) (số gạo)

Vậy cửa hàng còn lại số phần gạo là:

           \(1-\dfrac{3}{7}=\dfrac{4}{7}\) (số gạo)

                      Đáp số:.......số gạo

Cửa hàng còn lại :

`1-1/7-2/7=4/7` (số gạo ban đầu)

\(x+\dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{6}=-1\\ =>\dfrac{6x}{6}+\dfrac{3x}{6}+\dfrac{2x}{6}+\dfrac{x}{6}=-1\\ =>\dfrac{6x+3x+2x+x}{6}=-1\\ =>\dfrac{12x}{6}=-1\\ =>2x=-1\\ =>x=-\dfrac{1}{2}\)

\(x+\dfrac{x}{2}+\dfrac{x}{3}+\dfrac{x}{6}=-1\)

\(\dfrac{6x+3x+2x+x}{6}=-1\)

\(\dfrac{12x}{6}=-1\)

\(2x=-1\)

\(x=-\dfrac{1}{2}\)