1. Tìm x \(\in\) Q, biết
a) (x-\(\dfrac{1}{3}\))\(^2\) = 0
b) (x+\(\dfrac{1}{2}\))\(^2\) = \(\dfrac{1}{16}\)
c) (2x-1)\(^3\) = 8
Giúp mk vs
Những câu hỏi liên quan
Giải phương trình sau:
a. dfrac{6x+1}{x^2-7x+10}+dfrac{5}{x-2}dfrac{3}{x-5}
b. dfrac{2}{x^2-4}-dfrac{x-1}{xleft(x-2right)}+dfrac{x-4}{xleft(x+2right)}0
c. dfrac{1}{3-x}-dfrac{1}{x+1}dfrac{x}{x-3}-dfrac{left(x-1right)^2}{x^2-2x-3}
d. dfrac{1}{x-2}-dfrac{6}{x+3}dfrac{5}{6-x^2-x}
e. dfrac{2}{x+2}-dfrac{2x^2+16}{x^3+8}dfrac{5}{x^2-2x-3}
f. dfrac{x+1}{x^2+x+1}-dfrac{x-1}{x^2-x+1}dfrac{2left(x+2right)^2}{x^6-1}
m.n jup vs mai kt r
Đọc tiếp
Giải phương trình sau:
a. \(\dfrac{6x+1}{x^2-7x+10}+\dfrac{5}{x-2}=\dfrac{3}{x-5}\)
b. \(\dfrac{2}{x^2-4}-\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x-4}{x\left(x+2\right)}=0\)
c. \(\dfrac{1}{3-x}-\dfrac{1}{x+1}=\dfrac{x}{x-3}-\dfrac{\left(x-1\right)^2}{x^2-2x-3}\)
d. \(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)
e. \(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x-3}\)
f. \(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)
m.n jup vs mai kt r
Bài này t mới làm xong :v
d) \(\dfrac{1}{x-2}\) - \(\dfrac{6}{x+3}\) = \(\dfrac{5}{6-x^2-x}\) (ĐKXĐ : x \(\ne\) 2; x \(\ne\) -3)
<=> \(\dfrac{1}{x-2}\) - \(\dfrac{6}{x+3}\) = \(\dfrac{-5}{\left(x-2\right)\left(x+3\right)}\)
=> (x + 3) - 6(x - 2) = -5
<=> x + 3 - 6x + 12 = -5
<=> -5x + 15 = -5
<=> 5x = 20
<=> x = 4
Vậy tập nghiệm của PT là : S = { 4 }
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Tìm phân thức P biết :
a) \(p=\dfrac{4x^2-16}{2x+1}=\dfrac{4x^2+4x+1}{x-2}\)
b) \(\dfrac{2x^2+4x+8}{x^3-3x^2-x+3}:P=\dfrac{x^3-8}{\left(x+1\right)\left(x-3\right)}\)
a x^2-x0 b x^2-2x0 c (x+1)(x+2)(2-x)(x+2)
d dfrac{1}{x-2}+3dfrac{3-x}{x-2} đ dfrac{8-x}{x-7}-8dfrac{1}{x-7}
e dfrac{x}{2left(x-3right)}+dfrac{x}{2left(x+1right)}dfrac{2x}{left(x+1right)left(x-3right)}
f 5+dfrac{76}{x^2-16}dfrac{2x-1}{x+4}-dfrac{3x-1}{4-x}
g dfrac{90}{x}-dfrac{36}{x-6}2 h dfrac{1}{x}+dfrac{1}{x+10}dfrac{1}{12} i dfrac{x+3}{x-3}-dfrac{1}{...
Đọc tiếp
a \(x^2-x=0\) b \(x^2-2x=0\) c (x+1)(x+2)=(2-x)(x+2)
d \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\) đ \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
e \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)
f \(5+\dfrac{76}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)
g \(\dfrac{90}{x}-\dfrac{36}{x-6}=2\) h \(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\) i \(\dfrac{x+3}{x-3}-\dfrac{1}{x}=\dfrac{3}{x\left(x-3\right)}\)
k \(\dfrac{3}{x+2}-\dfrac{2}{x-2}+\dfrac{8}{x^2-4}=0\) l \(\dfrac{3}{x+2}-\dfrac{2}{x-3}=\dfrac{8}{\left(x-3\right)\left(x+2\right)}\)
m\(\dfrac{x}{2x+6}-\dfrac{x}{2x+2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
n \(\dfrac{x}{x+1}-\dfrac{2x-3}{1-x}=\dfrac{3x^2+5}{x^2-1}\) j \(\dfrac{5}{x+7}+\dfrac{8}{2x+14}=\dfrac{3}{2}\)
q \(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
Cần gấp ạ
câu nào cũng ghi lại đề nha
a) \(x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b)\(x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
c) \(\left(x+1\right)\left(x+2\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
d) \(\dfrac{1}{x-2}+3-\dfrac{3-x}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3\left(x-2\right)-\left(3-x\right)}{x-2}=0\)
\(\Leftrightarrow\dfrac{1+3x-6-3+x}{x-2}=0\) ( đk \(x\ne2\) )
\(\Leftrightarrow4x-8=0\Rightarrow x=2\)
đ) \(\dfrac{8-x}{x-7}-8-\dfrac{1}{x-7}=0\)
\(\Leftrightarrow\dfrac{8-x-8\left(x-7\right)-1}{x-7}=0\) (đk \(x\ne7\))
\(\Leftrightarrow8-x-8x+56-1=0\)
\(\Leftrightarrow-9x+63=0\)
\(\Leftrightarrow x=7\)
Đúng 0
Bình luận (0)
e) \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) (MTC 2(x+1)(x-3) )
\(\Leftrightarrow\dfrac{x\left(x+1\right)+x\left(x-3\right)-2x.2}{2\left(x-3\right)\left(x+1\right)}=0\) (đk \(x\ne3vsx\ne-1\) )
\(\Leftrightarrow x^2+x+x^2-3x-4x=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Tính:
\(a,\dfrac{x+3}{2x-1}-\dfrac{x^2-5}{4x^2-4x+1}-\dfrac{2x^3+5x^2-x-1}{8x^3-12x^2+6x-1}\)
\(b,\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
a: \(=\dfrac{4x^3+8x^2-11x+3-\left(x^2-5\right)\left(2x-1\right)-2x^3-5x^2+x+1}{\left(2x-1\right)^3}\)
\(=\dfrac{2x^3+3x^2-10x+4-2x^3+x^2+10x-5}{\left(2x-1\right)^3}\)
\(=\dfrac{4x^2-1}{\left(2x-1\right)^3}=\dfrac{2x+1}{\left(2x-1\right)^2}\)
b: \(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1+x^{32}}\)
Đúng 0
Bình luận (0)
Tính:
\(a,\dfrac{x+3}{2x-1}-\dfrac{x^2-5}{4x^2-4x+1}-\dfrac{2x^3+5x^2-x-1}{8x^3-12x^2+6x-1}\)
\(b,\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
GIẢI CÁC PHƯƠNG TRÌNH :
a). \(\dfrac{x^2-2x+2}{x^2-x+1}-\dfrac{x^2}{x^2+x+1}=\dfrac{3}{(x^4+x^2+1)x}\)
b).\(\dfrac{x^2+2x}{(x+1)^2+3}-\dfrac{x^2-2x}{(x-1)^2-3}=\dfrac{16}{x^4+4x^2+16}\)
GIÚP MÌNH VS CÁC BẠN!!!
a: \(\Leftrightarrow\dfrac{\left(x^2-2x+2\right)\left(x^2+x+1\right)-x^2\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(\Leftrightarrow\dfrac{x^4+x^3+x^2-2x^3-2x^2-2x+2x^2+2x+2-x^4+x^3-x^2}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
\(\Leftrightarrow x\cdot2=3\)
=>x=3/2
b:
Sửa đề; \(\dfrac{x^2+2x}{\left(x+1\right)^2+3}-\dfrac{x^2-2x}{\left(x-1\right)^2+3}=\dfrac{16}{x^4+4x^2+16}\)
\(\Leftrightarrow\dfrac{x^2+2x}{x^2+2x+4}-\dfrac{x^2-2x}{x^2-2x+4}=\dfrac{16}{x^4+4x^2+16}\)
\(\Leftrightarrow\left(x^2+2x\right)\left(x^2-2x+4\right)-\left(x^2-2x\right)\left(x^2+2x+4\right)=16\)
\(\Leftrightarrow x^4-2x^3+4x^2+2x^3-4x^2+8x-\left(x^4+2x^3+4x^2-2x^3-4x^2-8x\right)=16\)
=>\(x^4+8x-x^4+8x=16\)
=>16x=16
=>x=1
Đúng 0
Bình luận (0)
Tìm \(x\in\mathbb{Q}\), biết rằng :
a) \(\left(x-\dfrac{1}{2}\right)^2=0\)
b) \(\left(x-2\right)^2=1\)
c) \(\left(2x-1\right)^3=-8\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
a) \(\left(x-\dfrac{1}{2}\right)^2=0\Rightarrow x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
b) Vì \(\left(x-2\right)^2=1\Rightarrow\left\{{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
Vậy x = 4 hoặc x = 0
c) Vì \(\left(2.x-1\right)^3=-8\Rightarrow2.x-1=-2\Rightarrow2.x=-1\Rightarrow x=-\dfrac{1}{2}\)
d) Vì \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
a) \(\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
b) \(\left(x-2\right)^2=1\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) \(\left(2x-1\right)^3=-8\Leftrightarrow2x-1=-2\Leftrightarrow2x=-1\Leftrightarrow x=\dfrac{-1}{2}\) d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\Leftrightarrow x=\dfrac{-1}{4}\)
Đúng 0
Bình luận (0)
a: (x-1) ( x+5) > 0
b: (x-1) ( x+5) < 0
c) \(\left(x-\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
d)\(\dfrac{3}{4}-\dfrac{1}{4}\left|x-\dfrac{1}{7}\right|=\dfrac{1}{4}\)
e) 8 ( x+1) - 2 ( 2x + 5 ) = 0
g) ( 6x-1) - ( x + 8 ) =0
h) \(\left|7x-\dfrac{1}{4}\right|=1\)
x) -2x-3=-x+7
\(\left(x-1\right)\left(x+5\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\end{matrix}\right.\)
\(\left(x-1\right)\left(x+5\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1>0\Rightarrow x>1\\x+5< 0\Rightarrow x< -5\end{matrix}\right.\\\left\{{}\begin{matrix}x-1< 0\Rightarrow x< 1\\x+5>0\Rightarrow x>-5\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-5< x< 1\)
câu dễ tự làm
\(\Rightarrow x>-5;x< -5\)
Đúng 0
Bình luận (1)
Thực hiện các phép tính với các phân thức sau:
a) dfrac{4a^2-3a+5}{a^3-1}-dfrac{1-2a}{a^2+a+1}-dfrac{6}{a-1}
b) dfrac{5}{a+1}-dfrac{10}{a-left(a^2+1right)}-dfrac{15}{a^3+1}
c) dfrac{2x}{x^2+2xy}+dfrac{y}{xy-2y^2}+dfrac{4}{x^2-4y^2}
d) dfrac{1}{x-y}+dfrac{3xy}{y^3-x^3}+dfrac{x-y}{x^2+xy+y^2}
e) dfrac{2x+y}{2x^2-xy}+dfrac{16x}{y^2-4x^2}+dfrac{2x-y}{2x^2+xy}
f) dfrac{1}{1-x}+dfrac{1}{1+x}+dfrac{2}{1+x^2}+dfrac{4}{1+x^4}+dfrac{8}{1+x^8}+dfrac{16}{1+x^{16}}
Đọc tiếp
Thực hiện các phép tính với các phân thức sau:
a) \(\dfrac{4a^2-3a+5}{a^3-1}-\dfrac{1-2a}{a^2+a+1}-\dfrac{6}{a-1}\)
b) \(\dfrac{5}{a+1}-\dfrac{10}{a-\left(a^2+1\right)}-\dfrac{15}{a^3+1}\)
c) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
d) \(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\)
e) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
f) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
a: \(=\dfrac{4a^2-3a+5}{\left(a-1\right)\left(a^2+a+1\right)}+\dfrac{\left(2a-1\right)\left(a-1\right)}{\left(a-1\right)\left(a^2+a+1\right)}-\dfrac{6a^2+6a+1}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{4a^2-3a+5+2a^2-3a+1-6a^2-6a-6}{\left(a-1\right)\left(a^2+a+1\right)}\)
\(=\dfrac{-12a}{\left(a-1\right)\left(a^2+a+1\right)}\)
b: \(=\dfrac{5}{a+1}+\dfrac{10}{a^2-a+1}-\dfrac{15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2-5a+5+10a+10-15}{\left(a+1\right)\left(a^2-a+1\right)}\)
\(=\dfrac{5a^2+5a}{\left(a+1\right)\left(a^2-a+1\right)}=\dfrac{5a}{a^2-a+1}\)
Đúng 0
Bình luận (0)