Tìm x thuộc Q biết rằng
a) ( x- \(\frac{1}{2}\)) 2= 0
b) ( x-2) 2=1
c) ( 2x-1) 3= -8
d) ( x+ \(\frac{1}{2}\)) 2= \(\frac{1}{16}\)
a.(x-1/2)^2=0
b.(x-2)^2=1
c.(2x-1)^3=-8
d.(x+1/2)^2=1/6
a) \(\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
b) \(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
c) \(\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-3\Leftrightarrow x=-1\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{\sqrt{6}}{6}\\x+\dfrac{1}{2}=-\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{6}}{6}\\x=-\dfrac{3+\sqrt{6}}{6}\end{matrix}\right.\)
b: Ta có: \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c: Ta có: \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
hay \(x=-\dfrac{1}{2}\)
a. \(\left(x-\dfrac{1}{2}\right)^2=0\)
<=> \(x-\dfrac{1}{2}=0\)
<=> \(x=\dfrac{1}{2}\)
b. (x - 2)2 = 1
<=> (x - 2)2 - 12 = 0
<=> (x - 2 - 1)(x - 2 + 1) = 0
<=> (x - 3)(x - 1) = 0
<=> \(\left[{}\begin{matrix}x-3=0\\x-1=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c. (2x - 1)3 = -8
<=> (2x - 1)3 + 23 = 0
<=> (2x - 1 + 2)\(\left[\left(2x-1\right)^2-2\left(2x-1\right)+2^2\right]=0\)
<=> (2x + 1)(4x2 - 4x + 1 - 4x + 2 + 4) = 0
<=> (2x + 1)(4x2 - 8x + 7) = 0
<=> \(\left[{}\begin{matrix}2x+1=0\\4x^2-8x+7=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{-1}{2}\\S=\varnothing\end{matrix}\right.\)
d. \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{6}\)
<=> \(\left(x+\dfrac{1}{2}\right)^2-\dfrac{1}{\left(\sqrt{6}\right)^2}=0\)
<=> \(\left(x+\dfrac{1}{2}-\dfrac{1}{\sqrt{6}}\right)\left(x+\dfrac{1}{2}+\dfrac{1}{\sqrt{6}}\right)=0\)
<=> \(\left(x+\dfrac{3-\sqrt{6}}{6}\right)\left(x+\dfrac{3+\sqrt{6}}{6}\right)=0\)
<=> \(\left[{}\begin{matrix}x+\dfrac{3-\sqrt{6}}{6}=0\\x+\dfrac{3+\sqrt{6}}{6}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{6}}{6}\\x=\dfrac{-3-\sqrt{6}}{6}\end{matrix}\right.\)
tìm x thuộc Q, biết rằng :
a) \(\left(x-\frac{1}{2}\right)^2=0\)
b)\(\left(x-2\right)^2=1\)
c) \(\left(2x-1\right)^3=-8\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(x-\frac{1}{2}=0\)
\(x=0+\frac{1}{2}\)
\(x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\left(x-2\right)^2=1^2\)
\(x-2=1\)
\(x=1+2\)
\(x=3\)
c) \(\left(2x-1\right)^3=\left(-8\right)\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
\(2x-1=\left(-2\right)\)
\(2x=\left(-2\right)+1\)
\(2x=-1\)
\(x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\left(x+\frac{1}{2}\right)^2=\left(\frac{1}{4}\right)^2\)
\(x+\frac{1}{2}=\frac{1}{4}\)
\(x=\frac{1}{4}-\frac{1}{2}\)
\(x=-\frac{1}{4}\)
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
\(\Leftrightarrow x=\frac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=1\\x-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
c) \(\left(2x-1\right)^2=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=-\frac{1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}}\)
\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right):\left(\frac{x^2-2x}{x^3-x^2+x}\right)\))
a) Rút gọn
b) Tính giá trị A biết\(|x-\frac{3}{4}|=\frac{5}{4}\)
c) Tìm x thuộc Z để A thuộc Z
\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right)\div\left(\frac{x^2-2x}{x^3-x^2+x}\right)\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1}{x^2-x+1}-\frac{2}{x+1}\right)\div\left(\frac{x\left(x-2\right)}{x\left(x^2-x+1\right)}\right)\)
\(=\left(\frac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right)\div\frac{x-2}{x^2-x+1}\)
\(=\left(\frac{x+1+x+1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}\right)\times\frac{x^2-x+1}{x-2}\)
\(=\frac{-2x^2+4x}{\left(x+1\right)\left(x^2-x+1\right)}\times\frac{x^2-x+1}{x-2}\)
\(=\frac{-2x\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\frac{-2x}{x+1}\)
b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)
<=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-\frac{1}{2}\left(nhan\right)\end{cases}}\)
Với x = -1/2 => \(A=\frac{-2\cdot\left(-\frac{1}{2}\right)}{-\frac{1}{2}+1}=2\)
c) Để A ∈ Z thì \(\frac{-2x}{x+1}\)∈ Z
=> -2x ⋮ x + 1
=> -2x - 2 + 2 ⋮ x + 1
=> -2( x + 1 ) + 2 ⋮ x + 1
Vì -2( x + 1 ) ⋮ ( x + 1 )
=> 2 ⋮ x + 1
=> x + 1 ∈ Ư(2) = { ±1 ; ±2 }
x+1 | 1 | -1 | 2 | -2 |
x | 0 | -2 | 1 | -3 |
Các giá trị trên đều tm \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
Vậy x ∈ { -3 ; -2 ; 0 ; 1 }
Tìm x thuộc Q, biết rằng:
a) \(\left(x-\frac{1}{2}\right)^2=0\)
b) \(\left(x-2\right)^2=1\)
c) \(\left(2x-1\right)^3=-8\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
a) \(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{2}\end{cases}}\)
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left(x-2\right)^2-1=0\)
\(\Leftrightarrow\left(x-2-1\right)\left(x-2+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)
c) \(\left(2x-1\right)^3=-8\)
\(\Leftrightarrow2x-1=-2\)
\(\Leftrightarrow2x=-1\)
\(\Leftrightarrow x=\frac{-1}{2}\)
d) \(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\)
\(\Rightarrow x+\frac{1}{2}=\frac{1}{4}\)
\(\Rightarrow x=-\frac{1}{4}\)
a>(x-1/2)^2=0
(x-1/2)^2=0^2
=>x-1/2=0
x=1/2
b>(x-2)^2=1
(x-2)^2=1^2
=>x-2=1
x=3
(2x-1)^3=-8
=>(2x-1)^3=-2^3
=>2x-1=-2
2x=-1
x=-0,5
d>(x+1/2)^2=1/16
(x+1/2)^2=(1/4)^2
=>x+1/2=1/4
x=1/4-1/2
x=-1/4
chuc ban may man trong cuoc song!!!!!
1 thực hiện phép tính
a,\(\frac{x+3}{x+1}-\frac{2x-1}{x-1}-\frac{x-3}{x^2-1}\)
b, \(\frac{1}{x^2+x+1}+\frac{1}{x^2-x}+\frac{2x}{1-x^3}\)
c, \(\frac{1}{x-1}-\frac{1}{x+1}-\frac{2}{x^2+1}-\frac{4}{x^4+1}-\frac{8}{x^8+1}-\frac{16}{x^{16}+1}\)
giải phương trình hộ minh nha mấy bạn <3
a) \(\frac{3x-1}{x-1}-\frac{2x+5}{3}+\frac{4}{x^2-2x-3}=1\)
b) \(\frac{5}{x^2+x-6}+\frac{2}{x^2+4x+3}=\frac{-3}{2x-1}\)
c) \(\frac{4x^2+16}{x^2+16}=\frac{3}{x^2+1}+\frac{5}{x^2+3}+\frac{7}{x^2+5}\)
Làm đc 2 bài đầu chưa, t làm câu cuối cho, hai câu đầu dễ í mà
\(A=\left(\frac{x^2-2x}{2x^2+8}-\frac{2x^2}{8-4x+2x^2-x^3}\right)\left(1-\frac{1}{x}-\frac{2}{x^2}\right)\)
Tìm x thuộc Z để A thuộc Z
Bài 1: Tìm x,y:
a) |x - 1| + |x + 3| = 4
b) |2x + 3| + |2x - 1| = \(\frac{8}{2\left(y-5\right)^2+2}\)
c) |x + 3| + |x + 1| = \(\frac{16}{\left|y-2\right|+\left|y+2\right|}\)
Bài 2: Tìm số nguyên x,y, biết:
a) \(\frac{1}{x}+\frac{1}{y}=\frac{1}{5}\)
b) \(x^2-2xy+y=0\)
a)Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:
\(\left|x-1\right|+\left|3+x\right|=\left|1-x\right|+\left|3+x\right|\ge\left|1-x+3+x\right|=4\)
\(\Rightarrow VT\ge VP."="\Leftrightarrow-3\le x\le1\)
b) \(\hept{\begin{cases}\left|2x+3\right|+\left|2x-1\right|=\left|2x+3\right|+\left|1-2x\right|\ge4\\\frac{8}{2\left(y-5\right)^2+2}\le4\end{cases}}\Leftrightarrow VT\ge VP."="\Leftrightarrow\hept{\begin{cases}-\frac{3}{2}\le x\le\frac{1}{2}\\y=5\end{cases}}\)
c Tương tự b
2) \(\frac{1}{x}+\frac{1}{y}=5\Leftrightarrow x+y-5xy=0\Leftrightarrow5x+5y-25xy=0\Leftrightarrow5x\left(1-5y\right)-\left(1-5y\right)=-1\)
\(\Leftrightarrow\left(5x-1\right)\left(1-5y\right)=-1\)
Xét ước
8,Thực hiện phép tính
a,\(\frac{5x^2-y^2}{xy}-\frac{3x-2y}{y}\)
b,\(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
c,\(\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}\)
d,\(\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}\)
e,\(\frac{2x+y}{2x^2-xy}+\frac{16x}{y^2-4x^2}+\frac{2x-y}{2x^2+xy}\)
f,\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)