Bài này có người giải rồi nhé, đừng đăng đi đăng lại ạ, từ qua giờ là 3 lần rồi c.

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

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

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)

\(-2=\frac{-2(x^3-1)}{x^3-1}\)

tìm các số nguyên x để mỗi phân số sau đây là số nguyên

Ngô Hải Nam ơi bn trả lời giúp mik ik

bài đó là bài 4^* tìm các số nguyên x để mỗi phân số sau đây là số nguyên

\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

x = nhân ạ

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\)

\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\times\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{4}{4}-\dfrac{1}{4}\right)\times\left(\dfrac{5}{5}-\dfrac{1}{5}\right)\)

\(=\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\)

\(=\dfrac{1\times2\times3\times4}{2\times3\times4\times5}\)

\(=\dfrac{1}{5}\)

\(\dfrac{1}{4}\cdot\dfrac{1}{3}+\dfrac{1}{6}\cdot\dfrac{1}{4}+\dfrac{1}{4}\cdot\dfrac{1}{7}\\ =\dfrac{1}{4}\cdot\left(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{7}\right)\\ =\dfrac{1}{4}\cdot\dfrac{9}{14}\\ =\dfrac{9}{56}\)

đoạn cuối thiếu dấu"+"

\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)

\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)

\(A=\sqrt{36}-\sqrt{4}=6-2=4\)

\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+...+\dfrac{1}{\sqrt{34}+\sqrt{35}}+\dfrac{1}{\sqrt{35}+\sqrt{36}}\)

\(=-\sqrt{4}+\sqrt{5}-\sqrt{5}+\sqrt{6}-...-\sqrt{35}+\sqrt{36}\)

\(=6-2=4\)

\(\dfrac{x-3}{7}=\dfrac{2x-7}{16}\)

⇒\(16\left(x-3\right)=7\left(2x-7\right)\)

⇒\(16x-48=14x-49\)

⇒\(16x-14x=-49+48\)

⇒\(2x=-1\)

⇒\(x=\dfrac{-1}{2}\)

Vậy \(x=\dfrac{-1}{2}\)

mik cảm ơn nha

a)

\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)

b)

\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)

c)

\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

d)

\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

e)

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)

f)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)

1) \(3^x=\dfrac{9^8}{27^3\cdot81^2}\)

\(\Rightarrow3^x=\dfrac{\left(3^2\right)^8}{\left(3^3\right)^3\cdot\left(3^4\right)^2}\)

\(\Rightarrow3^x=\dfrac{3^{16}}{3^{15}}\)

\(\Rightarrow3^x=3\)

\(\Rightarrow x=1\)

2) \(\dfrac{2^{4-x}}{16^5}=32^6\)

\(\Rightarrow\dfrac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

\(\Rightarrow\dfrac{2^{4-x}}{2^{20}}=2^{30}\)

\(\Rightarrow2^{4-x}=2^{20}\cdot2^{30}\)

\(\Rightarrow2^{4-x}=2^{50}\)

\(\Rightarrow4-x=50\)

\(\Rightarrow x=-46\)

3) \(\dfrac{2^{2x-3}}{4^{10}}=8^3\cdot16^5\)

\(\Rightarrow\dfrac{2^{2x-3}}{\left(2^2\right)^{10}}=\left(2^3\right)^3\cdot\left(2^4\right)^5\)

\(\Rightarrow\dfrac{2^{2x-3}}{2^{20}}=2^{29}\)

\(\Rightarrow2^{2x-3}=2^{49}\)

\(\Rightarrow2x-3=49\)

\(\Rightarrow2x=52\)

\(\Rightarrow x=26\)