A²=b(a-c)-c(a-b)

=ab-bc-ac+bc=ab-ac=a.(b-c)=-20.(-5)=100

=>A=+100

a) \(\left(3-x\right)\left(5x+10\right)=0\)

\(\left(3-x\right).5.\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3-x=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

vậy...

b) \(\left|5x+2\right|-4x=7\)

\(\left|5x+2\right|=7+4x\)

\(\Rightarrow\orbr{\begin{cases}5x+2=7+4x\\5x+2=-7-4x\end{cases}}\Rightarrow\orbr{\begin{cases}5x-4x=7-2\\5x+4x=-7-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\9x=-9\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

vậy.... 

k nha 

a, \(\left(3-x\right)\left(5x+10\right)=0\)

\(\Rightarrow3-x=0\)   hoặc     \(5x+10=0\)

\(\Rightarrow x=3\)         hoặc          \(x=-2\)

\(\left|5x+2\right|-4x=7\)

\(\Rightarrow\left|5x+2\right|=7+4x\)

\(\Rightarrow\orbr{\begin{cases}5x+2=7+4x\\5x+2=-\left(7+4x\right)\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}5x-4x=7-2\\5x+2=-7-4x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\5x+4x=-7-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\9x=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

a) \(\frac{\sqrt{640}\sqrt{34,3}}{\sqrt{567}}\)

\(= \frac{\sqrt{64.10}\sqrt{49.\frac{7}{10}}}{\sqrt{81.7}}\)

\(= \frac{\sqrt{64}\sqrt{10}\sqrt{49}\sqrt{\frac{7}{10}}}{\sqrt{81}\sqrt{7}}\)

\(= \frac{\sqrt{64}\sqrt{49}}{\sqrt{81}} . \frac{\sqrt{10}\sqrt{\frac{7}{10}}}{\sqrt{7}}\)

\(= \frac{8.7}{9} . \frac{\sqrt{10 . \frac{7}{10}}}{\sqrt{7}}\)

\(= \frac{56}{9} . \frac{\sqrt{7}}{\sqrt{7}}\)

\(= \frac{56}{9} . 1 = \frac{56}{9}\)

b) \(\sqrt{21,6}\sqrt{810}\sqrt{11^2−5^2}\)

\(= \sqrt{216.\frac{1}{10}}\sqrt{81.10}\sqrt{(11−5)(11+5)}\)

\(= \sqrt{36.6.\frac{1}{10}}\sqrt{81}\sqrt{10}\sqrt{6.16}\)

\(= \sqrt{36}\sqrt{6}\sqrt{\frac{1}{10}}\sqrt{81}\sqrt{10}\sqrt{6}\sqrt{16}\)

\(= (\sqrt{36}\sqrt{81}\sqrt{16}).(\sqrt{6}\sqrt{6}).(\sqrt{\frac{1}{10}}\sqrt{10})\)

\(= (6.9.4).\sqrt{6.6}.\sqrt{\frac{1}{10}.10}\)

\(= (54.4).\sqrt{36}.\sqrt{1}\)

\(= 216.6.1 = 1296\)

a) Vì 12 ⋮ 3x + 1 => 3x + 1 ∊ Ư(12) = {-12;-6;-4;-3;-2;-1;1;2;3;4;6;12} => 3x ∊ {-13;-7;-5;-4;-3;-2;0;1;2;3;5;11}. Vì 3x ⋮ 3 => 3x ∊ {-3;0;3} => x ∊ {-1;0;1}. Vậy x ∊ {-1;0;1}. b) 2x + 3 ⋮ 7 => 2x + 3 ∊ B(7) = {...;-21;-14;-7;0;7;14;21;...}. Vì 2x ⋮ 2 mà 3 lẻ nên khi số lẻ trừ đi 3 thì 2x mới ⋮ 2 => 2x + 3 lẻ => 2x + 3 ∊ {...;-35;-21;-7;7;21;35;...} => 2x ∊ {...;-38;-24;-10;4;18;32;...} => x ∊ {...;-19;-12;-5;2;9;16;...} => x ⋮ 7 dư 2 => x = 7k + 2. Vậy x = 7k + 2 (k ∊ Z)

Cảm ơn bạn nhiều nha Đặng Hoàng Lâm!

hi mk sẽ chia sẻ câu hỏi này giúp bn

uhm, cảm ơn nha

Pikachu đơn giản thì làm thử đừng nói mà ko làm nha ^_^

duyệt đi

Đơn giản -_-

\(\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{101}{1504}\)

(=)\(\frac{1}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{101}{1540}\)

(=)\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+..+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

(=)\(\frac{1}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{101}{1540}\)

(=)\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{5}{1540}=\frac{1}{308}\)

\(\Rightarrow\)x=305

a) \(\left(x-7\right)-\left(5-x\right)=12-\left(-8+x\right)\)

\(\Leftrightarrow x-7-5+x=12+8-x\)

\(\Leftrightarrow2x-12=20-x\)

\(\Leftrightarrow2x+x=12+20\)

\(\Leftrightarrow3x=32\)

\(\Leftrightarrow x=\frac{32}{3}\)

b) 124 + ( 13 - 16 ) = 162 - ( x + 162 )

<=> 124 - 3 = 162 - x - 162

<=> 121 = -x

<=> x = -121

Ix-7I+24=24-x+(-24)+x+40

Ix-7I+24=24+(-24)+40+(-x+x)

Ix-7I+24=40

Ix-7I=40-24

Ix-7I=16

x-7=16                             hoặc   x-7=-16

x=16+7                                      x=-16+7

x=23                                          x=-9

Vậy x=23 hoặc x=-9

Ix-7I+24=24-x+(-24)+x+40

Ix-7I+24=24+(-24)+40+(-x+x)

Ix-7I+24=40

Ix-7I=40-24

Ix-7I=16

x-7=16                             hoặc   x-7=-16

x=16+7                                      x=-16+7

x=23                                          x=-9

Vậy x=23 hoặc x=-9