a) \(x^2-36=0\)

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

\(\Leftrightarrow\left(x-6\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\) 

Vậy: ... 

b) \(x^2-10x+25=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot5+5^2=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy: ... 

a) \(x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy \(x\in\left\{6;-6\right\}\)

b) \(x^2-10x+25=0\)

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

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

Vậy \(x=5\)

(2x - 5) + 17 = 6

2x - 5 = 6 - 17 

2x - 5 = - 11

2x = -11 + 5

2x = -6

x = `(-6)/2` 

x = -3

(lớp 5 chưa học số âm) 

( 2x - 5 ) + 17 = 6

2x - 5 = 6 - 17

2x - 5 = -11

2x = -11 + 5

2x = -6

x = -6 : 2

x = -3

a) \(\left(3x-1\right)\left(x+2\right)-\left(x+2\right)^2\)

\(=\left(3x^2+6x-x-2\right)-\left(x+2\right)^2\)

\(=\left(3x^2+5x-2\right)-\left(x^2+4x+4\right)\)

\(=3x^2+5x-2-x^2-4x-4\)

\(=2x^2+x-6\) 

b) \(\left(x-1\right)\left(x+1\right)-\left(x^2-2x+1\right)\)

\(=\left(x^2-1\right)-\left(x^2-2x+1\right)\) 

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

\(=2x-2\)

c) \(\left(x-4\right)\left(4+x\right)+2x\left(x-3\right)\)

\(=\left(x-4\right)\left(x+4\right)+2x\left(x-3\right)\)

\(=\left(x^2-16\right)+2x^2-6x\)

\(=x^2-16+2x^2-6x\)

\(=3x^2-6x-16\)

d) \(\left(x-1\right)\left(x^2-1\right)+\left(x+2\right)^3\)

\(=\left(x^3-x-x^2+1\right)+\left(x^3+6x^2+12x+8\right)\)

\(=x^3-x-x^2+1+x^3+6x^2+12x+8\)

\(=2x^3+5x^2+11x+9\)

e) \(\left(2x-1\right)^2-\left(2x-5\right)\left(x+5\right)\)

\(=\left(4x^2-4x+1\right)-\left(2x^2+10x-5x-25\right)\)

\(=\left(4x^2-4x+1\right)-\left(2x^2+5x-25\right)\)

\(=4x^2-4x+1-2x^2-5x+25\)

\(=2x^2-9x+26\)

f) \(\left(3x+1\right)^2-\left(x^2-1\right)\left(x^2+2\right)\)

\(=\left(9x^2+6x+1\right)-\left(x^4+2x^2-x^2-2\right)\)

\(=\left(9x^2+6x+1\right)-\left(x^4+x^2-2\right)\)

\(=9x^2+6x+1-x^4-x^2+2\)

\(=-x^4+8x^2+6x+3\) 

g) \(\left(x^2+1\right)^2-\left(x^2-1\right)\left(x^2+2\right)\)

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

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

\(=x^4+2x^2+1-x^4-x^2+2\)

\(=x^2+3\)

h) \(\left(2x^2-4\right)^2-\left(2x^2+4\right)^2\)

\(=\left(4x^4-16x^2+16\right)-\left(4x^4+16x^2+16\right)\)

\(=4x^4-16x^2+16-4x^4-16x^2-16\)

\(=-32x^2\)

a) Nữa chu vi mảnh đất là:

90 : 2 = 45 (m) 

Tổng số phần bằng nhau là:

2 + 3 = 5 (phần)

Chiều rộng là:

45 : 5 x 2 = 18 (m)

Chiều dài là:

45 - 18 = 27 (m)

Diện tích mảnh đất là:

18 x 27 = 486 `(m^2)` 

b) Diện tích trồng rau là:

20% x 486 = 97,2 `(m^2)` 

Diện tích trồng hoa là:

`2/9 xx 486 = 108 (m^2)` 

Diện tích trồng cây ăn quả là:

486 - 97,2 - 108 = 280,8 `(m^2)`

ĐS: ... 

gấp ạa

\(\dfrac{34+17\times14}{16\times36-32}\)

\(=\dfrac{34+\left(17\times2\right)\times7}{16\times36-16\times2}\)

\(=\dfrac{34+34\times7}{16\times\left(36-2\right)}\)

\(=\dfrac{34\times\left(1+7\right)}{34\times16}\)

\(=\dfrac{8}{16}\)

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

2B. 

a) \(A=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}\cdot\dfrac{\dfrac{3}{4}\cdot\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}\cdot\dfrac{3}{4}+\dfrac{5}{8}\)

\(=\dfrac{3}{8}+\dfrac{5}{8}\)

\(=\dfrac{8}{8}\)

\(=1\)

b) \(B=\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)

\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}-\dfrac{1}{7}}{\dfrac{3}{8}-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{5}}{\dfrac{3}{4}+\dfrac{1}{2}-\dfrac{3}{10}}\)

\(=\dfrac{\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}}{3\left(\dfrac{1}{8}-\dfrac{1}{5}+\dfrac{1}{7}\right)}+\dfrac{2\cdot\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}{\dfrac{3}{4}+\dfrac{3}{6}-\dfrac{3}{10}}\)

\(=\dfrac{1}{3}\cdot\dfrac{2\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}{3\left(\dfrac{1}{4}+\dfrac{1}{6}-\dfrac{1}{10}\right)}\)

\(=\dfrac{1}{3}\cdot\dfrac{2}{3}\)

\(=\dfrac{2}{9}\)

3A:

\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}=\dfrac{-1}{10}>-\dfrac{1}{9}\)

3B:

\(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\cdot\left(\dfrac{1}{2}+1\right)\cdot\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{11}{10}\)

\(=\dfrac{-1}{10}\cdot\dfrac{11}{2}=\dfrac{-11}{20}\)

Vì 20<21 nên \(\dfrac{11}{20}>\dfrac{11}{21}\)

=>\(-\dfrac{11}{20}< -\dfrac{11}{21}\)

=>\(B< -\dfrac{11}{21}\)

\(2,5-3x=5,5.2022^0\)

\(=>2,5-3x=5,5.1\)

\(=>2,5-3x=5,5\)

\(=>3x=2,5-5,5\)

\(=>3x=-3\)

\(=>x=\left(-3\right):3\)

\(=>x=\dfrac{-3}{3}=-1\)

Vậy...

\(#NqHahh\)

\(2,5-3x=5,5\cdot2022^0\)

\(2,5-3x=5,5\cdot1\)

\(2,5-3x=5,5\)

\(3x=2,5-5,5\)

\(3x=-3\)

\(x=-3:3\)

\(x=-1\)

Vậy \(x=-1\)

Xét ΔABC vuông tại A có \(\widehat{B}+\widehat{C}=90^0\)

nên \(sinB=cosC=\dfrac{4}{5}\)

\(sin^2B+cos^2B=1\)

=>\(cos^2B=1-\left(\dfrac{4}{5}\right)^2=\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\)

=>\(cosB=\dfrac{3}{5}\)

\(tanB=\dfrac{sinB}{cosB}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)

\(cotB=\dfrac{1}{tanB}=\dfrac{3}{4}\)

Vì tam giác ABC vuông tại A

Nên: \(\widehat{B}+\widehat{C}=90^o\\ \Rightarrow0^o< \widehat{C}< 90^o\)

\(\Rightarrow0< \sin C< 1\) 

Ta có: \(\sin^2C+\cos^2C=1\Rightarrow\sin^2C=1-\left(\dfrac{4}{5}\right)^2=\dfrac{9}{25}\\ \Rightarrow\sin C=\dfrac{3}{5}\)

Lại có: \(\tan C=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{4}\\ \cot C=\dfrac{1}{\tan C}=\dfrac{4}{3}\)