Bài 2: Tìm x để các phân thức sau bằng 0
a, \(\dfrac{2x-1}{x^2-1}\)
b,\(\dfrac{\left(x+2\right)\left(x-3\right)}{2x-5}\)
c,\(\dfrac{x^2+x}{2x+1}\)
Tìm điều kiện của x để phân thức sau xác định;
a)\(\dfrac{\dfrac{1}{x-4}}{2x+2}\)
b)\(\dfrac{x^3+2x}{4x^2-25}\)
c)\(\dfrac{2x^2+2x}{8x^3+27}\)
d)\(\dfrac{2x+1}{\left(2x+2\right)\left(4y^2-9\right)}\)
`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`
`<=>x ne 4,x me -1`
`b,ĐKXĐ:4x^2-25 ne 0`
`<=>(2x-5)(2x+5) ne 0`
`<=>x ne +-5/2`
`c,ĐKXĐ:8x^3+27 ne 0`
`<=>8x^3 ne -27`
`<=>2x ne -3`
`<=>x ne -3/2`
`d,2x+2 ne 0,4y^2-9 ne 0`
`<=>2x ne -2,(2y-3)(2y+3) ne 0`
`<=>x ne -1,y ne +-3/2`
b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)
d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)
Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
Bài `1`: Rút gọn các biểu thức sau:
\(a)4x^2\left(5x^2+3\right)-6x\left(3x^3-2x+1\right)-5x^3\left(2x-1\right)\)
\(b)\dfrac{3}{2}x\left(x^2-\dfrac{2}{3}x+2\right)-\dfrac{5}{3}x^2\left(x+\dfrac{6}{5}\right)\)
Bài `2`: Thực hiện các phép nhân sau:
\(a)\left(x^2-x\right)\cdot\left(2x^2-x-10\right)\)
\(b)\left(0,2x^2-3x\right)\cdot5\left(x^2-7x+3\right)\)
\(c)6x^2\cdot\left(2x^3-3x^2+5x-4\right)\)
\(d)\left(-1,2x^2\right)\cdot\left(2,5x^4-2x^3+x^2-1,5\right)\)
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
tìm x để các biểu thức sau có nghĩa:
a)\(\sqrt{\left(x-2\right)}\)+\(\dfrac{1}{x-5}\) b)\(\sqrt{\left(2x-6\right)\left(7-x\right)}\) c)\(\sqrt{4x^2-25}\)
d)\(\dfrac{2}{x^2-9}\)-\(\sqrt{5-2x}\) e)\(\dfrac{x}{x^2-4}\)+\(\sqrt{x-2}\)
a) \(\sqrt{x-2}+\dfrac{1}{x-5}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ge0\\x-5\ne0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge2\\x\ne5\end{matrix}\right.\)
b) \(\sqrt{\left(2x-6\right)\left(7-x\right)}=\sqrt{2\left(x-3\right)\left(7-x\right)}\) có nghĩa khi:
\(\left(x-3\right)\left(7-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\7-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\7-x\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge3\\x\le7\end{matrix}\right.\\\left\{{}\begin{matrix}x\le3\\x\ge7\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow3\le x\le7\)
c) \(\sqrt{4x^2-25}=\sqrt{\left(2x-5\right)\left(2x+5\right)}\) có nghĩa khi:
\(\left(2x-5\right)\left(2x+5\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-5\ge0\\2x+5\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-5\le0\\2x+5\le0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge\dfrac{5}{2}\\x\ge-\dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{2}\\x\le-\dfrac{5}{2}\end{matrix}\right.\)
d) \(\dfrac{2}{x^2-9}-\sqrt{5-2x}=\dfrac{2}{\left(x+3\right)\left(x-3\right)}-\sqrt{5-2x}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\5-2x\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\le\dfrac{5}{2}\end{matrix}\right.\)
e) \(\dfrac{x}{x^2-4}+\sqrt{x-2}=\dfrac{x}{\left(x+2\right)\left(x-2\right)}+\sqrt{x-2}\) có nghĩa khi:
\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm2\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow x>2\)
Bài 1. Giải các bất phương trình sau 1) \(\dfrac{2x-1}{x+1}-2< 0\) 2) \(\dfrac{x^2-2x+5}{x-2}-x+1\ge0\)
3) \(\dfrac{\left(1+2x\right)\left(x-3\right)}{\left(2x+3\right)\left(1-x\right)}\le0\) 4) \(\left|2x-3\right|>5\) 5)\(\left|1-2x\right|\le4\)
6) \(\left|3x+1\right|>x-2\)
\(\dfrac{2x-1}{x+1}-2< 0.\left(x\ne-1\right).\\ \Leftrightarrow\dfrac{2x-1-2x-2}{x+1}< 0.\Leftrightarrow\dfrac{-3}{x+1}< 0.\)
Mà \(-3< 0.\)
\(\Rightarrow x+1>0.\Leftrightarrow x>-1\left(TMĐK\right).\)
\(\dfrac{x^2-2x+5}{x-2}-x+1\ge0.\left(x\ne2\right).\\ \Leftrightarrow\dfrac{x^2-2x+5-x^2+2x+x-2}{x-2}\ge0.\\ \Leftrightarrow\dfrac{x+3}{x-2}\ge0.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3\ge0.\\x-2\ge0.\end{matrix}\right.\\\left\{{}\begin{matrix}x+3\le0.\\x-2\le0.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge-3.\\x\ge2.\end{matrix}\right.\\\left\{{}\begin{matrix}x\le-3.\\x\le2.\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x\ge2.\\x\le-3.\end{matrix}\right.\)
Kết hợp ĐKXĐ.
\(\Rightarrow\left[{}\begin{matrix}x>2.\\x\le-3.\end{matrix}\right.\)
\(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}\le0.\left(x\ne1;x\ne\dfrac{-3}{2}\right).\)
Đặt \(\dfrac{\left(1+2x\right)\left(x-2\right)}{\left(2x+3\right)\left(1-x\right)}=f\left(x\right).\)
Ta có bảng sau:
\(x\) | \(-\infty\) \(-\dfrac{3}{2}\) \(-\dfrac{1}{2}\) \(1\) \(2\) \(+\infty\) |
\(1+2x\) | - | - 0 + | + | + |
\(x-2\) | - | - | - | - 0 + |
\(2x+3\) | - 0 + | + | + | + |
\(1-x\) | + | + | + 0 - | - |
\(f\left(x\right)\) | - || + 0 - || + 0 - |
Vậy \(f\left(x\right)\ge0.\Leftrightarrow x\in\left(\dfrac{-3}{2};\dfrac{-1}{2}\right)\cup\)(1;2].
Rút gọn, rồi tính giá trị các phân thức sau : A=\(\dfrac{\left(2x^{2^{ }}+2x^{ }\right)\left(x-2\right)^2}{^{ }\left(x^{3^{ }}-4x\right)\left(x+1\right)}\)với x = \(\dfrac{1}{2}\)
B=\(\dfrac{x^3-x^{2^{ }}y+xy^2}{x^3+y^3}\)với x = -5 , y = 10
\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)
\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)
Bài 1: Giải hệ phương trình sau
\(\left\{{}\begin{matrix}\dfrac{1}{2x-y}+\left(x+3y\right)=\dfrac{3}{2}\\\dfrac{4}{2x-y}-5\left(x+3y\right)=-2\end{matrix}\right.\)
Bài 2: Cho phương trình: x\(^2\)+(m-1)x-m\(^2\)-2=0
a) CMR: phương trình luôn có 2 nghiệm phân biệt \(\forall\)m
b) Tìm m để biểu thức A=\(\left(\dfrac{x_1}{x_2}\right)^3+\left(\dfrac{x_2}{x_1}\right)^3\) đạt giá trị lớn nhất.
Bài 2:
a) Ta có: \(\Delta=\left(m-1\right)^2-4\cdot1\cdot\left(-m^2-2\right)\)
\(=m^2-2m+1+4m^2+8\)
\(=5m^2-2m+9>0\forall m\)
Do đó, phương trình luôn có hai nghiệm phân biệt với mọi m
Bài 1:
ĐKXĐ \(2x\ne y\)
Đặt \(\dfrac{1}{2x-y}=a;x+3y=b\)
HPT trở thành
\(\left\{{}\begin{matrix}a+b=\dfrac{3}{2}\\4a-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\4\left(\dfrac{3}{2}-b\right)-5b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{3}{2}-b\\6-9b=-2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{8}{9}\\a=\dfrac{11}{18}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3y=\dfrac{8}{9}\\2x-y=\dfrac{18}{11}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=2x-\dfrac{18}{11}\\x+3\left(2x-\dfrac{18}{11}\right)=\dfrac{8}{9}\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{82}{99}\\y=\dfrac{2}{99}\end{matrix}\right.\)
Tìm x để biểu thức sau có gt dương:
\(A=x^2-3x\)
Tìm x để các biểu thức sau có gía trị âm:
\(D=x^2+\dfrac{5}{2}x\\ E=\dfrac{x-3}{x-2}\\ G=\left(2x-1\right)\left(3-2x\right)\)
a: A>0
=>\(x^2-3x>0\)
=>x(x-3)>0
TH1: \(\left\{{}\begin{matrix}x>0\\x-3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>0\\x>3\end{matrix}\right.\)
=>x>3
TH2: \(\left\{{}\begin{matrix}x< 0\\x-3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 0\\x< 3\end{matrix}\right.\)
=>x<0
d: Để D<0 thì \(x^2+\dfrac{5}{2}x< 0\)
=>\(x\left(x+\dfrac{5}{2}\right)< 0\)
TH1: \(\left\{{}\begin{matrix}x>0\\x+\dfrac{5}{2}< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>0\\x< -\dfrac{5}{2}\end{matrix}\right.\)
=>Loại
Th2: \(\left\{{}\begin{matrix}x< 0\\x+\dfrac{5}{2}>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 0\\x>-\dfrac{5}{2}\end{matrix}\right.\)
=>\(-\dfrac{5}{2}< x< 0\)
e: ĐKXĐ: x<>2
Để E<0 thì \(\dfrac{x-3}{x-2}< 0\)
TH1: \(\left\{{}\begin{matrix}x-3>=0\\x-2< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>=3\\x< 2\end{matrix}\right.\)
=>Loại
TH2: \(\left\{{}\begin{matrix}x-3< =0\\x-2>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< =3\\x>2\end{matrix}\right.\)
=>2<x<=3
g: Để G<0 thì \(\left(2x-1\right)\left(3-2x\right)< 0\)
=>\(\left(2x-1\right)\left(2x-3\right)>0\)
TH1: \(\left\{{}\begin{matrix}2x-1>0\\2x-3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>\dfrac{3}{2}\end{matrix}\right.\)
=>\(x>\dfrac{3}{2}\)
TH2: \(\left\{{}\begin{matrix}2x-1< 0\\2x-3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x< \dfrac{3}{2}\end{matrix}\right.\)
=>\(x< \dfrac{1}{2}\)
Rút gọn các phân thức sau:
a) \(\dfrac{6x^2y^2}{8xy^{ }5}\)
b) \(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
c) \(\dfrac{2x^2+2x
}{x+1}\)
d) \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
e) \(\dfrac{36\left(x-2\right)^3}{32-16x}\)
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)