\(\dfrac{74-x}{26}+\dfrac{75-x}{25}\)
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9 tháng 2 2021
Sửa đề: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)Ta có: \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
mà \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}>0\)
nên 100-x=0
hay x=100
Vậy: S={100}
9 tháng 2 2021
Ta có : \(\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}=-5\)
\(\Leftrightarrow\dfrac{74-x}{26}+\dfrac{75-x}{25}+\dfrac{76-x}{24}+\dfrac{77-x}{23}+\dfrac{78-x}{22}+5=0\)
\(\Leftrightarrow\dfrac{74-x}{26}+1+\dfrac{75-x}{25}+1+\dfrac{76-x}{24}+1+\dfrac{77-x}{23}+1+\dfrac{78-x}{22}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{26}+\dfrac{100-x}{25}+\dfrac{100-x}{24}+\dfrac{100-x}{23}+\dfrac{100-x}{22}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\right)=0\)
Thấy : \(\dfrac{1}{26}+\dfrac{1}{25}+\dfrac{1}{24}+\dfrac{1}{23}+\dfrac{1}{22}\ne0\)
\(\Rightarrow100-x=0\)
\(\Leftrightarrow x=100\)
Vậy ...
AH
Akai Haruma
Giáo viên
23 tháng 8 2023
Lời giải:
PT $\Leftrightarrow \frac{x+25}{75}+1+\frac{x+30}{70}+1=\frac{x+35}{65}+1+\frac{x+40}{60}+1$
$\Leftrightarrow \frac{x+100}{75}+\frac{x+100}{70}=\frac{x+100}{65}+\frac{x+100}{60}$
$\Leftrightarrow (x+100)(\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60})=0$
Dễ thấy $\frac{1}{75}+\frac{1}{70}-\frac{1}{65}-\frac{1}{60}<0$
$\Rightarrow x+100=0$
$\Leftrightarrow x=-100$ (tm)
28 tháng 1 2022
\(\dfrac{x^2-26}{10}+\dfrac{x^2-25}{11}\ge\dfrac{x^2-24}{12}+\dfrac{x^2-23}{13}\)
\(\Leftrightarrow\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)
\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)
\(\Leftrightarrow\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)
\(\Leftrightarrow\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)
Vì \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\Rightarrow x^2-36\ge0\Leftrightarrow\left[{}\begin{matrix}x\le-6\\x\ge6\end{matrix}\right.\)
28 tháng 1 2022
Bất phương trình đó tương đương với:
\(\left(\dfrac{x^2-26}{10}-1\right)+\left(\dfrac{x^2-25}{11}-1\right)\ge\left(\dfrac{x^2-24}{12}-1\right)+\left(\dfrac{x^2-23}{13}-1\right)\)
⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}\ge\dfrac{x^2-36}{12}+\dfrac{x^2-36}{13}\)
⇔ \(\dfrac{x^2-36}{10}+\dfrac{x^2-36}{11}-\dfrac{x^2-36}{12}-\dfrac{x^2-36}{13}\ge0\)
⇔ \(\left(x^2-36\right)\left(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}\right)\ge0\)
+)Vì \(\dfrac{1}{10}>\dfrac{1}{11}>\dfrac{1}{12}>\dfrac{1}{13}\) nên \(\dfrac{1}{10}+\dfrac{1}{11}-\dfrac{1}{12}-\dfrac{1}{13}>0\)
⇔ \(x^2-36\ge0\)
⇔ \(x^2\ge36\)
⇔ \(\sqrt{x^2}\ge6\)
⇔ \(\left|x\right|\ge6\)
⇔ \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)
➤ Vậy \(\left[{}\begin{matrix}x\ge6\\x\le-6\end{matrix}\right.\)
26 tháng 1 2019
Mỗi số hạng của vế trái cộng thêm 1, vế phải = 5. Mỗi số hạng vế trái có mẫu số giống nhau, bạn đặt x+ 2020 làm nhân tử chung, phần còn lại tự làm nhé.
mấy bài còn lại bạn đăng cx làm tương tự
27 tháng 1 2019
\(\frac{x+24}{1996}+\frac{x+25}{1995}+\frac{x+26}{1994}+\frac{x+27}{1993}+\frac{x+2036}{4}=0\)
\(\Leftrightarrow\left(\frac{x+24}{1996}+1\right)+\left(\frac{x+25}{1995}+1\right)+\left(\frac{x+26}{1994}+1\right)+\left(\frac{x+27}{1993}+1\right)+\left(\frac{x+2036}{4}-4\right)=0\)
\(\Leftrightarrow\frac{x+2020}{1996}+\frac{x+2020}{1995}+\frac{x+2020}{1994}+\frac{x+2020}{1993}+\frac{x+2020}{4}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{1996}+\frac{1}{1995}+\frac{1}{1994}+\frac{1}{1993}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy ....
1 tháng 8 2021
\(\dfrac{x-12}{77}+\dfrac{x-11}{78}=\dfrac{x-74}{15}+\dfrac{x-73}{16}\)
\(\Rightarrow\dfrac{x-12}{77}+\dfrac{x-11}{78}-\dfrac{x-74}{15}-\dfrac{x-73}{16}=0\)
\(\Rightarrow\dfrac{x-12}{77}-1+\dfrac{x-11}{78}-1-\dfrac{x-74}{15}+1-\dfrac{x-73}{16}+1=0+1+1-1-1\)
\(\Rightarrow\left(\dfrac{x-12}{77}-1\right)+\left(\dfrac{x-11}{78}-1\right)-\left(\dfrac{x-74}{15}-1\right)-\left(\dfrac{x-73}{16}-1\right)=0\)
\(\Rightarrow\dfrac{x-89}{77}+\dfrac{x-89}{78}-\dfrac{x-89}{15}-\dfrac{x-89}{16}=0\)
\(\Rightarrow\left(x-89\right)\left(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-89=0\\\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}=0\end{matrix}\right.\)
\(x-89=0\\ \Rightarrow x=89\)
\(\dfrac{1}{77}+\dfrac{1}{78}-\dfrac{1}{15}-\dfrac{1}{16}=0\)(vô lí)
Vậy \(x=89\)
HP
15 tháng 12 2020
Đặt \(x+\dfrac{1}{x}=a;y+\dfrac{1}{y}=b\left(\left|a\right|\ge2;\left|b\right|\ge2\right)\)
\(\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\x^3+y^3+\dfrac{1}{x^3}+\dfrac{1}{y^3}=15m-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x^3+\dfrac{1}{x^3}\right)+\left(y^3+\dfrac{1}{y^3}\right)=15m-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)+\left(y+\dfrac{1}{y}\right)^3-3\left(y+\dfrac{1}{y}\right)=15m-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3-3\left(x+\dfrac{1}{x}+y+\dfrac{1}{y}\right)=15m-25\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=5\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3=15m-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\a^3+b^3=15m-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\\left(a+b\right)^3-3ab\left(a+b\right)=15m-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\125-15ab=15m-10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=5\\ab=9-m\end{matrix}\right.\)
\(\Rightarrow a,b\) là nghiệm của phương trình \(t^2-5t+9-m=0\left(1\right)\)
a, Nếu \(m=3\), phương trình \(\left(1\right)\) trở thành
\(t^2-5t+6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\\\left\{{}\begin{matrix}a=3\\b=2\end{matrix}\right.\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\\y+\dfrac{1}{y}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\y^2-3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{3\pm\sqrt{5}}{2}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+\dfrac{1}{x}=3\\y+\dfrac{1}{y}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3\pm\sqrt{5}}{2}\\y=1\end{matrix}\right.\)
Vậy ...
b, \(\left(1\right)\Leftrightarrow t=\dfrac{5\pm\sqrt{4m-11}}{2}\left(m\ge\dfrac{11}{4}\right)\)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{5\pm\sqrt{4m-11}}{2}\\b=\dfrac{5\mp\sqrt{4m-11}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=\dfrac{5\pm\sqrt{4m-11}}{2}\\y+\dfrac{1}{y}=\dfrac{5\mp\sqrt{4m-11}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x^2-\left(5\pm\sqrt{4m-11}\right)+2=0\left(2\right)\\2y^2-\left(5\mp\sqrt{4m-11}\right)+2=0\end{matrix}\right.\)
Yêu cầu bài toán thỏa mãn khi phương trình \(\left(2\right)\) có nghiệm dương
\(\Leftrightarrow\left\{{}\begin{matrix}\Delta=\left(5\pm\sqrt{4m-11}\right)^2-16\ge0\\\dfrac{5\pm\sqrt{4m-11}}{2}>0\\1>0\end{matrix}\right.\)
\(\Leftrightarrow...\)
11 tháng 3 2021
\(PT\Leftrightarrow\left(\dfrac{x-70}{130}-1\right)+\left(\dfrac{x-25}{175}-1\right)+\left(\dfrac{x-50}{150}-1\right)+\left(\dfrac{x-275}{25}+3\right)=0\)
\(\Leftrightarrow\left(x-200\right)\left(\dfrac{1}{130}+\dfrac{1}{175}+\dfrac{1}{150}+\dfrac{1}{25}\right)=0\Leftrightarrow x=200\).
Vậy...
PP
1
18 tháng 3 2023
\(\Leftrightarrow\dfrac{16}{x+4}+\dfrac{16}{x-4}=\dfrac{5}{3}\)
=>\(\dfrac{16x-64+16x+64}{x^2-16}=\dfrac{5}{3}\)
=>5(x^2-16)=3*32x=96x
=>5x^2-96x-80=0
=>x=20 hoặc x=-4/5
PP
Phùng Phạm Quỳnh Trang
CTVHS
19 tháng 3 2023
nếu là giải PT bằng cách quy đồng:
25x2 + 480 - 400 = 0
làm sao để phan tích ra ạ.
20 tháng 3 2023
1: =>2(x+2)>3x+1
=>2x+4-3x-1>0
=>-x+3>0
=>-x>-3
=>x<3
2: =>12x^2-2x>12x^2+9x-8x-6
=>-2x>-x-6
=>-x>-6
=>x<6
3: =>4(x+1)-12>=3(x-2)
=>4x+4-12>=3x-6
=>4x-8>=3x-6
=>x>=2
4: =>-5x<=15
=>x>=-3
5: =>3(x+2)-5(x-2)<30
=>3x+6-5x+10<30
=>-2x+16<30
=>-2x<14
=>x>-7
6: =>5(x+2)<3(3-2x)
=>5x+10<9-6x
=>11x<-1
=>x<-1/11
\(\dfrac{74-x}{26}=\dfrac{75-x}{25}\\ \Leftrightarrow\dfrac{74-x}{26}+1=\dfrac{75-x}{25}+1\\ \Leftrightarrow\dfrac{74-x}{26}+\dfrac{26}{26}=\dfrac{75-x}{25}+\dfrac{25}{25}\\ \Leftrightarrow\dfrac{100-x}{26}=\dfrac{100-x}{25}\\ \Leftrightarrow\dfrac{100-x}{26}-\dfrac{100-x}{25}=0\\ \Leftrightarrow\left(100-x\right)\left(\dfrac{1}{26}-\dfrac{1}{25}\right)=0\)
\(Do:\dfrac{1}{26}-\dfrac{1}{25}\ne0\\ \Rightarrow100-x=0\Rightarrow x=100 \)
Vậy S = { 100 }