`@` `\text {Ans}`

`\downarrow`

`a)`

`13/50 + 9% + 41/100 + 0,24`

`= 0,26 + 0,09 + 0,41 + 0,24`

`= (0,26 + 0,24) + (0,09 + 0,41)`

`= 0,5 + 0,5`

`= 1`

`b)`

`2018 \times 2020 - 1/2017 + 2018 \times 2019`

`= 2018 \times (2020 + 2019) - 1/2017`

`= 2018 \times 4039 - 1/2017`

`= 8150702`

`c)`

`1/2 + 1/6 + 1/12 + 1/20 +1/30 +1/42`

`=`\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}\)

`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{6}-\dfrac{1}{7}\)

`=`\(1-\dfrac{1}{7}\)

`= 6/7`

\(a,\dfrac{13}{50}+9\%+\dfrac{41}{100}+0,24\\ 0,26+0,09+0,41+0,24\\ =\left(0,26+0,24\right)+\left(0,09+0,41\right)\\ =0,5+0,5\\ =1\\ b,2018\times2020-\dfrac{1}{2017}+2018\times2019\\ =2018\times\left(2020+2019\right)-\dfrac{1}{2017}\\ =2018\times4039-\dfrac{1}{2017}\\ =3150702-\dfrac{1}{2017}\\ c,\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\\ =1-\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}.........+\dfrac{1}{6}-\dfrac{1}{7}\\ =1-\dfrac{1}{7}\\ =\dfrac{6}{7}\)

Cảm ơn các bạn đã trả lời giúp mình câu hỏi nha

Sửa đề: x² + 13x + 41 --> x² + 13x + 42

Giải:

\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+41}=\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{2}\)

(ĐKXĐ: \(x\ne\left\{-1;-2;-3;-4;-5;-6;-7\right\}\))

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{2}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+7}=\frac{1}{2}\)

\(\Leftrightarrow\frac{x+7-x-1}{\left(x+1\right)\left(x+7\right)}=\frac{1}{2}\)

\(\Leftrightarrow\left(x+1\right)\left(x+7\right)=12\)

\(\Leftrightarrow x^2+8x+7=12\)

⇔$x^2-8x=5$

⇔ $8x+{(-4)}^2=5+{(-4)}^2$
⇔ $x^2-8x+16=21$
⇔ ${(x-4)}^2=21$
⇔ $x=\pm \sqrt{21}+4$

Vậy...

Chúc bạn học tốt@@

vabh ơi cho mk hỏi bạn có ghi sai đề k ạ?

sr mik viết sai đề 42 ko phải 41

Đk:\(x\ne2;x\ne3;x\ne4;x\ne5;x\ne6\)

\(pt\Leftrightarrow\frac{1}{\left(x-6\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-4\right)}+...+\frac{1}{\left(x-3\right)\left(x-2\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-5}+\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-4}+...+\frac{1}{x-3}-\frac{1}{x-2}=\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{x-6}-\frac{1}{x-2}=\frac{1}{8}\)\(\Leftrightarrow\frac{x-2}{\left(x-6\right)\left(x-2\right)}-\frac{x-6}{\left(x-2\right)\left(x-6\right)}=\frac{1}{8}\)

\(\Leftrightarrow\frac{4}{\left(x-6\right)\left(x-2\right)}=\frac{1}{8}\Leftrightarrow\left(x-2\right)\left(x-6\right)=32\)

\(\Leftrightarrow x^2-8x+12=32\Leftrightarrow x^2-8x-20=0\)

\(\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=10\\x=-2\end{cases}}\)

! là gì vậy bn sao ghi vào?

giải phuong trinh a bn

\(\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}=\frac{1}{x+2}-\frac{1}{\left(x+6\right)}\)

\(\frac{1}{t}-\frac{1}{t+4}=\frac{4}{t\left(t+4\right)}=\frac{1}{8}=\frac{4}{32}\Rightarrow t=4\Rightarrow x=2\)

\(\dfrac{1}{x^2-5x+6}+\dfrac{1}{x^2-7x+12}+\dfrac{1}{x^2-9x+20}+\dfrac{1}{x^2-11x+30}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-\dfrac{1}{x-4}+\dfrac{1}{x-4}-\dfrac{1}{x-5}+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{1}{x-6}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{x-6-x+2}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-2\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow32=\left(x-2\right)\left(x-6\right)\)

\(\Leftrightarrow32=x^2-8x+12\)

\(\Leftrightarrow x^2+8x-20=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=10\end{matrix}\right.\)

1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅

3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1

5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)

6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅

7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅

8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1

9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)

\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)

\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

Câu 3, 4 tương tự nhé.

\(5.\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Leftrightarrow\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-5-1\right)\left(x-5+1\right)=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-6\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x=6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

Vậy \(x\in\left\{4;5;6\right\}\)

Mình ko hiểu cái bài mẫu, mình chỉ viết đáp án thui được ko? Nhưng bạn check xem đáp án mình đúng ko nhé.

1/3 gấp 4 lần 1/12. 1/4 gấp 3 lần 1/12. 1/6 gấp 2 lần 1/12.

Hok tốt

Ừm bn cho thế này cũng đc . Lại còn đúng

a)\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{10}+x=\frac{3}{5}\)

\(\Rightarrow\frac{2}{5}+x=\frac{3}{5}\)

\(\Rightarrow x=\frac{3}{5}-\frac{2}{5}=\frac{1}{5}\)

b)\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+...+\frac{2}{13}-\frac{2}{15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{2}{3}-\frac{2}{15}+x=\frac{1}{3}\)

\(\Rightarrow\frac{8}{15}+x=\frac{1}{3}\)

\(\Rightarrow x=\frac{1}{3}-\frac{8}{15}=-\frac{1}{5}\)

c)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Rightarrow\frac{1}{1}-\frac{1}{x+1}=\frac{9}{10}\)

\(\Leftrightarrow\frac{x+1-1}{x+1}=\frac{9}{10}\)

\(\Rightarrow\frac{x}{x+1}=\frac{9}{10}\)

\(\Rightarrow x=9\)

b) \(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{15-13}{13.15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}+x=\frac{1}{3}\)

\(\Leftrightarrow\frac{1}{3}-\frac{1}{15}+x=\frac{1}{3}\)

\(\Leftrightarrow x=\frac{1}{15}\)