Tìm x biết: \(\frac{x+2}{x-1}=\frac{x-3}{x+1}\)
Giải hộ mk nnha ~
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Ta có: \(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}-2\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{x}\\y=\frac{1}{y}\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x^2=1\\y^2=1\end{cases}}\)
\(\Leftrightarrow\)(x, y) = (1, 1; 1, - 1; - 1, 1; - 1, - 1)
- rut gon di ban
- sau rut gon A se co dang a/b
- theo yeu cau thi bieu thuc co gia tri am tuc la : a va b trai dau
a>0, b< 0
a<0, b> 0
~~^^~~
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{8}-\frac{2}{5}x=\frac{17}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{2}{5}x=\frac{17}{4}+\frac{3}{8}\)(Bạn tự quy đồng chỗ này)
\(\Leftrightarrow\frac{1}{10}x=\frac{37}{8}\)
\(\Leftrightarrow x=\frac{185}{4}\)
Ta có :
\(\frac{x-1}{49}+\frac{x-2}{48}+\frac{x-3}{47}+\frac{x-4}{46}+\frac{x-5}{45}=5\)
\(\Leftrightarrow\)\(\left(\frac{x-1}{49}-1\right)+\left(\frac{x-2}{48}-1\right)+\left(\frac{x-3}{47}-1\right)+\left(\frac{x-4}{46}-1\right)+\left(\frac{x-5}{45}-1\right)=5-5\)
\(\Leftrightarrow\)\(\frac{x-1-49}{49}+\frac{x-2-48}{48}+\frac{x-3-47}{47}+\frac{x-4-46}{46}+\frac{x-5-45}{45}=0\)
\(\Leftrightarrow\)\(\frac{x-50}{49}+\frac{x-50}{48}+\frac{x-50}{47}+\frac{x-50}{46}+\frac{x-50}{45}=0\)
\(\Leftrightarrow\)\(\left(x-50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\) ( vì nó lớn hơn 0 )
Nên \(x-50=0\)
\(\Rightarrow\)\(x=50\)
Vậy \(x=50\)
Chúc bạn học tốt ~
\(2,\)
\(a,\sqrt{x^2-4x+3}=3\)
\(\Rightarrow x^2-4x+3=9\)
\(\Rightarrow x^2-4x-6=0\)
\(\Rightarrow\left(x-2\right)^2=10\)
\(\Rightarrow\orbr{\begin{cases}x-2=\sqrt{10}\\x-2=-\sqrt{10}\end{cases}\Rightarrow\orbr{\begin{cases}x=2+\sqrt{10}\\x=2-\sqrt{10}\end{cases}}}\)
Ta có:
\(x+\frac{1}{x}=5\)
\(\Rightarrow\frac{2x+1}{x}=\frac{5x}{x}\)
\(\Rightarrow2x+1-5x=0\)
\(\Rightarrow-3x=-1\Rightarrow x=\frac{1}{3}\)
Thay x=1/3 vào \(x^2+\frac{1}{x^2}=\left(\frac{1}{3}\right)^2+\frac{1}{\left(\frac{1}{3}\right)^2}=\frac{1}{9}+1:\frac{1}{9}=\frac{82}{9}\)
Tương tự với \(x^3+\frac{1}{x^3}\)
a, (x2 - 5)(x2 - 24) < 0
=> x2 - 5 và x2 - 24 trái dấu
Mà x2 - 5 > x2 - 24 => \(\hept{\begin{cases}x^2-5>0\\x^2-24>0\end{cases}\Rightarrow5< x^2< 24}\)
Vì x \(\in\)Z nên x2 = 9;16
+) x2 = 9 => x = 3 hoặc x = -3
+) x2 = 16 => x = 4 hoặc x = -4
Vậy...
b,
\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
\(\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà \(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)\ne0\)
=> x + 1 = 0 => x = 0 - 1 => x = -1
\(\frac{x+1}{14}+\frac{x+2}{13}=\frac{x+3}{12}+\frac{x+4}{11}\)
\(\Rightarrow\left(\frac{x+1}{14}+1\right)+\left(\frac{x+2}{13}+1\right)=\left(\frac{x+3}{12}+1\right)+\left(\frac{x+4}{11}+1\right)\)
\(\Rightarrow\frac{x+15}{14}+\frac{x+15}{13}=\frac{x+15}{12}+\frac{x+15}{11}\)
\(\Rightarrow\frac{x+15}{14}+\frac{x+15}{13}-\frac{x+15}{12}-\frac{x+15}{11}=0\)
\(\Rightarrow\left(x+15\right)\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)=0\)
Mà \(\left(\frac{1}{14}+\frac{1}{13}-\frac{1}{12}-\frac{1}{11}\right)\ne0\)
=> x + 15 = 0 => x = 0 - 15 => x = -15
\(\frac{x+2}{x-1}=\frac{x-3}{x+1}\Leftrightarrow\left(x+2\right)\left(x+1\right)=\left(x-1\right)\left(x-3\right)\Leftrightarrow x^2+3x+2=x^2-4x+3\)
\(\Leftrightarrow x^2+3x+2-x^2+4x-3=0\Leftrightarrow7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)
=>(x+2)(x+1)=(x-1)(x-3)
x.x+x.1+2x+2.1=x.x-3x-1x+1.3
x2+x+2x+2=x2-3x-1x+3
x2+(x+2x)+2=x2+(-3x-1x)+3
x2+3x+2=x2-4x+3
x2+3x-x2+4x=3-2
(x2-x2)+(3x+4x)=1
7x=1
x=1/7