Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
b) \(\frac{x-5}{2017}+\frac{x-2}{2020}=\frac{x-6}{2016}+\frac{x-68}{1954}\)
\(\Leftrightarrow\)\(\frac{x-5}{2017}-1+\frac{x-2}{2020}-1=\frac{x-6}{2016}-1+\frac{x-68}{1954}-1\)
\(\Leftrightarrow\)\(\frac{x-2022}{2017}+\frac{x-2022}{2020}=\frac{x-2022}{2016}+\frac{x-2022}{1954}\)
\(\Leftrightarrow\)\(\left(x-2022\right)\left(\frac{1}{2017}+\frac{1}{2020}-\frac{1}{2016}-\frac{1}{1954}\right)=0\)
\(\Leftrightarrow\)\(x-2022=0\) (vì 1/2017 + 1/2020 - 1/2016 - 1/1954 \(\ne0\))
\(\Leftrightarrow\)\(x=2022\)
Vậy...
b) \(\frac{x-5}{2017}+\frac{x-2}{2020}=\frac{x-6}{2016}+\frac{x-68}{1954}\)
\(\Leftrightarrow\)\(\frac{x-5}{2017}-1+\frac{x-2}{2020}-1=\frac{x-6}{2016}-1+\frac{x-68}{1954}-1\)
\(\Leftrightarrow\)\(\frac{x-2022}{2017}+\frac{x-2022}{2020}=\frac{x-2022}{2016}+\frac{x-2022}{1954}\)
\(\Leftrightarrow\)\(\left(x-2022\right)\left(\frac{1}{2017}+\frac{1}{2020}-\frac{1}{2016}-\frac{1}{1954}\right)=0\)
\(\Leftrightarrow\)\(x-2022=0\) (vì 1/2017 + 1/2020 - 1/2016 - 1/1954 \(\ne0\))
\(\Leftrightarrow\)\(x=2022\)
Vậy,....
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
a/ ĐKXĐ: x khác 1; x khác - 2
pt <=> \(\dfrac{x-1}{\left(x+2\right)\left(x-1\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-1\right)}=\dfrac{4x-8}{\left(x+2\right)\left(x-1\right)}\)
\(\Leftrightarrow x-1-2x-4=4x-8\Leftrightarrow-5x=-3\Leftrightarrow x=\dfrac{3}{5}\left(tm\right)\)
Vậy........
b/ \(2x-3\ge5\Leftrightarrow2x\ge8\Leftrightarrow x\ge4\)
Vậy......
c,d tt
a. \(\dfrac{1}{x+2}-\dfrac{2}{x-1}=\dfrac{4x-8}{\left(x+2\right)\left(x-1\right)}\)
ĐKXĐ: \(x\ne-2;x\ne1\)
\(\Leftrightarrow\dfrac{1\left(x-1\right)}{x+2\left(x-1\right)}-\dfrac{2\left(x+2\right)}{x-1\left(x+2\right)}=\dfrac{4x-8}{\left(x+2\right)\left(x-1\right)}\)
\(\Rightarrow1\left(x-1\right)-2\left(x+2\right)=4x-8\)
\(\Leftrightarrow x-1-2x-4=4x-8\)
\(\Leftrightarrow x-2x-4x=-8+1+4\)
\(\Leftrightarrow-5x=-3\)
\(\Leftrightarrow x=\dfrac{3}{5}\)
Vậy \(S=\left\{\dfrac{3}{5}\right\}\)
b) \(2x-3\ge5\left(2\right)\)
\(\Leftrightarrow2x\ge8\)
\(\Leftrightarrow x\ge4\)
Vậy tập nghiệm của BPT (2) là \(x\ge4\)
c) \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
ĐKXĐ: \(x\ne2;x\ne-1\)
\(\Leftrightarrow\dfrac{2\left(x-2\right)}{x+1\left(x-2\right)}-\dfrac{1\left(x+1\right)}{x-2\left(x+1\right)}=\dfrac{2x-6}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow2x-3-1-x=2x-6\)
\(\Leftrightarrow2x-x-2x=-6+3+1\)
\(\Leftrightarrow x=2\) (KTM)
Vậy pt vô \(n_o\)
d) \(3x-5\ge7\left(4\right)\)
\(\Leftrightarrow3x\ge12\)
\(\Leftrightarrow x\ge4\)
Vậy tập nghiệm của BPT (4) là \(x\ge4\)
a) 4x -8 ≥ 3(3x-1)-2x +1
⇒4x -8 ≥7x -2
⇒4x -7x ≥ -2 +8
⇒-3x ≥ 6
⇒x≤-2
Vậy bpt có nghiệm là:{x|x≤-2}
b) (x-3)(x+2)+(x+4)2≤ 2x (x+5)+4
⇔ x2+2x - 3x - 6 +x2 + 8x +16≤ 2x2 + 10x +4
⇔ x2 +2x - 3x + x2 + 8x - 2x2- 10x ≤ 4+6-16
⇔ -3x ≤ -6
⇔ x≥ 2
Vậy bpt có tập nghiệm là: {x|x≥2}
a. \(\dfrac{6x+5}{2}-\dfrac{10x+3}{4}=2x+\dfrac{2x+1}{2}\)
\(\Leftrightarrow2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)
\(\Leftrightarrow12x+10-10x-3=8x+4x+2\)
\(\Leftrightarrow12x-10x-8x-4x=2-10+3\)
\(\Leftrightarrow-10x=-5\Leftrightarrow x=\dfrac{1}{2}\)
b. \(\left(x+1\right)^3-\left(x-1\right)^3=6\left(x^2+x+1\right)\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1=6x^2+6x+6\)
\(\Leftrightarrow6x^2+2=6x^2+6x+6\)
\(\Leftrightarrow6x^2-6x^2-6x=6-2\Leftrightarrow-6x=4\)
\(\Leftrightarrow x=\dfrac{-2}{3}\)
c. \(\dfrac{x+2}{13}+\dfrac{2x+45}{15}=\dfrac{3x+8}{37}+\dfrac{4x+69}{9}\)
\(\Leftrightarrow\left(\dfrac{x+2}{13}+1\right)+\left(\dfrac{2x+45}{15}-1\right)=\left(\dfrac{3x+8}{37}+1\right)+\left(\dfrac{4x+69}{9}-1\right)\)
\(\Leftrightarrow\dfrac{x+15}{13}+\dfrac{2\left(x+15\right)}{15}-\dfrac{3\left(x+15\right)}{37}-\dfrac{4\left(x+15\right)}{9}=0\)
\(\Leftrightarrow\left(x+15\right)\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)=0\)
Vì \(\left(\dfrac{1}{13}+\dfrac{2}{15}-\dfrac{3}{37}-\dfrac{4}{9}\right)>0\)
\(\Leftrightarrow x+15=0\Leftrightarrow x=-15\)
a) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2=0\)
\(\Rightarrow\left(x^2+4x+8\right)^2+2.\dfrac{3}{2}x\left(x^2+4x+8\right)+\dfrac{9}{4}x^2-\dfrac{1}{4}x^2=0\)
\(\Rightarrow\left(x^2+4x+8+\dfrac{3}{2}x\right)^2-\left(\dfrac{1}{2}x\right)^2=0\)
\(\Rightarrow\left(x^2+4x+8+\dfrac{3}{2}x-\dfrac{1}{2}x\right)\left(x^2+4x+8+\dfrac{3}{2}x+\dfrac{1}{2}x\right)=0\)
\(\Rightarrow\left(x^2+4x+8+x\right)\left(x^2+4x+8+2x\right)=0\)
\(\Rightarrow\left(x^2+5x+8\right)\left(x^2+6x+8\right)=0\)
\(\Rightarrow\left(x^2+5x+8\right)\left(x^2+2x+4x+8\right)=0\)
\(\Rightarrow\left(x^2+5x+8\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]=0\)
\(\Rightarrow\left(x^2+5x+8\right)\left(x+2\right)\left(x+4\right)=0\)
Vì x2 ≥ 0 với mọi x
⇒ x2 + 5x + 8 ≥ 0 với mọi x
\(\Rightarrow\left(x+2\right)\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)
b) \(\dfrac{x-5}{2017}+\dfrac{x-2}{2020}=\dfrac{x-6}{2016}+\dfrac{x-68}{1954}\)
Trừ 2 vào mỗi vế ta có:
\(\Rightarrow\dfrac{x-5}{2017}-1+\dfrac{x-2}{2020}-1=\dfrac{x-6}{2016}-1+\dfrac{x-68}{1954}-1\)
\(\Rightarrow\dfrac{x-2022}{2017}+\dfrac{x-2022}{2020}-\dfrac{x-2022}{2016}-\dfrac{x-2022}{1954}=0\)
\(\Rightarrow\left(x-2022\right)\left(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\right)=0\)
Ta thấy \(\dfrac{1}{2017}+\dfrac{1}{2020}-\dfrac{1}{2016}-\dfrac{1}{1954}\ne0\)
\(\Rightarrow x-2022=0\Rightarrow x=2022\)
Chúc bạn học tốt!