c) \(\dfrac{y^4-1}{y^3+y^2+y+1}=\)
d)\(\dfrac{2x^2-9x+7}{-2x^2-x+28}=\)
Những câu hỏi liên quan
tìm điều kiện bài toán:
a) \(y=\dfrac{1}{x}-\dfrac{\sqrt{2x-1}}{x^2-3x+2}\)
b) \(y=\dfrac{1}{x^2-1}-\sqrt{7-2x}\)
c) \(y=\dfrac{2}{x}+\dfrac{3}{4-2x+x^2}\)
d) \(y=\sqrt{25-x^2}-2\sqrt{x}+3\)
Lời giải:
a.
\(\left\{\begin{matrix} x\neq 0\\ 2x-1\geq 0\\ x^2-3x+2=(x-1)(x-2)\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ x\geq \frac{1}{2}\\ x\neq 1; x\neq 2\end{matrix}\right.\)
$\Leftrightarrow x\geq \frac{1}{2}; x\neq 1; x\neq 2$
b. \(\left\{\begin{matrix}
x^2-1=(x-1)(x+1)\neq 0\\
7-2x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}
x\neq \pm 1\\
x\leq \frac{7}{2}\end{matrix}\right.\)
c.
\(\left\{\begin{matrix} x\neq 0\\ 4-2x+x^2\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 0\\ (x-1)^2+3\neq 0\end{matrix}\right.\Leftrightarrow x\neq 0\)
d.
\(\left\{\begin{matrix} 25-x^2=(5-x)(5+x)\geq 0\\ x\geq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} -5\leq x\leq 5\\ x\geq 0\end{matrix}\right.\Leftrightarrow 0\leq x\leq 5\)
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a) \(y=\dfrac{1}{x}-\dfrac{\sqrt[]{2x-1}}{x^2-3x+2}\)
Điều kiện \(\) \(2x-1\ge0;x\ne0;x^2-3x+2\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;\left(x-1\right)\left(x-2\right)\ne0\)
\(\Leftrightarrow x\ge\dfrac{1}{2};x\ne0;x\ne1;x\ne2\)
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a) \(x\ge\dfrac{1}{2};x\ne1;x\ne2\)
b) \(x\le\dfrac{7}{2};x\ne\pm1\)
c) \(x\ne0\)
d) \(0\le x\le5\)
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Giải các hệ phương trình sau
f.{ (2x - y) (x + 3y) = 4
{ (5x + y) (x + 3y) = 24
g.{ \(\dfrac{8x-5y-3}{7}+\dfrac{11y-4x-7}{5}=12\)
{ \(\dfrac{9x+4y-13}{5}+\dfrac{3\left(x-2\right)}{4}=15\)
h.{\(\dfrac{1}{x}+\dfrac{1}{y}=2\)
{\(\dfrac{3}{x}-\dfrac{4}{y}=-1\)
h) \(\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{1}{y}=2\\\dfrac{3}{x}-\dfrac{4}{y}=-1\end{matrix}\right.\)\(\left(1\right)\)\(\left(đk:x,y\ne0\right)\)
Đặt \(a=\dfrac{1}{x},b=\dfrac{1}{y}\)
\(\left(1\right)\Leftrightarrow\) \(\left\{{}\begin{matrix}a+b=2\\3a-4b=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3a+3b=6\\3a-4b=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\7b=7\end{matrix}\right.\)\(\Leftrightarrow a=b=1\)
Thay a,b:
\(\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{y}=1\Leftrightarrow x=y=1\left(tm\right)\)
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A dfrac{5xy^2-3z}{3xy}+dfrac{4x^2y+3z}{3xy}B dfrac{3y+5}{y-1}+dfrac{-y^2-4y}{1-y}+dfrac{y^2+y+7}{y-1}C dfrac{6x}{x^2-9}+dfrac{5x}{x-3}+dfrac{x}{x+3}D dfrac{1-3x}{2x}+dfrac{3x-2}{2x-1}+dfrac{3x-2}{2x-4x^2}E dfrac{x^3+2x}{x^3+1}+dfrac{2x}{x^2-x+1}+dfrac{1}{x+1}
Đọc tiếp
A = \(\dfrac{5xy^2-3z}{3xy}+\dfrac{4x^2y+3z}{3xy}\)
B = \(\dfrac{3y+5}{y-1}+\dfrac{-y^2-4y}{1-y}+\dfrac{y^2+y+7}{y-1}\)
C = \(\dfrac{6x}{x^2-9}+\dfrac{5x}{x-3}+\dfrac{x}{x+3}\)
D = \(\dfrac{1-3x}{2x}+\dfrac{3x-2}{2x-1}+\dfrac{3x-2}{2x-4x^2}\)
E = \(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}+\dfrac{1}{x+1}\)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
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Tìm x , y ,z :
a, \(\dfrac{x+z+1}{x}=\dfrac{z+x+2}{y}=\dfrac{x+y-3}{z}=\dfrac{1}{x+y+z}\)
b, 10x = 6y và \(2x^2-y^2=-28\)
c, \(\dfrac{1+2y}{18}=\dfrac{1+4y}{24}=\dfrac{1+6y}{6x}\)
d, \(\dfrac{2x+1}{5}=\dfrac{3y-2}{7}=\dfrac{2x+3y-1}{6x}\)
Ta có : 2x+1 /5 = 3y-2/7 = 2x+3y -1 /6x
=> 2x+1+3y-2 / 5+7 = 2x+3y-1 /6x
=> 2x+3y-1 / 12 = 2x+3y-1 / 6x
=> 12 = 6x => x =2
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Chứng minh các phương trình sai có vô số nghiệm
a,\(\left(2x-1\right)^2+\left(x+4\right)^2+5=2x\left(x+1\right)+\left(x+2\right)^2+2x^2-2x+18\)
b,\(\dfrac{5x-7}{4}-\dfrac{9x-4}{5}=-x-\dfrac{19-9x}{20}\)
c,\(|y-3|=y-3\)
a ) \(\left(2x-1\right)^2+\left(x+4\right)^2+5=2x\left(x+1\right)+\left(x+2\right)^2+2x^2-2x+18\)
\(\Leftrightarrow4x^2-4x+1+x^2+8x+16+5=2x^2+2x+x^2+4x+4+2x^2-2x+18\)
\(\Leftrightarrow5x^2+4x+22=5x^2+4x+22\)
=> PT có vô số nghiệm
b ) \(\dfrac{5x-7}{4}-\dfrac{9x-4}{5}=-x-\dfrac{19-9x}{20}\)
\(\Leftrightarrow\dfrac{25x-35-36x+16}{20}=\dfrac{-20x-19+9x}{20}\)
\(\Leftrightarrow\dfrac{-11x-19}{20}=\dfrac{-11x-19}{20}\)
=> PT có vô số nghiệm
c ) \(\left|y-3\right|=y-3\)
TH 1 : \(y\ge3\)
\(\Rightarrow y-3\ge0\Rightarrow\left|y-3\right|=y-3\)
Do \(\left|y-3\right|=y-3\)
\(\Rightarrow y-3=y-3\)
Nên : \(y\ge3\) , PT vô số nghiệm
TH 2 : \(y< 3\Rightarrow y-3< 0\Rightarrow\left|y-3\right|=3-y\)
Do \(\left|y-3\right|=y-3\)
\(\Rightarrow3-y=y-3\)
\(\Rightarrow3-y-y+3=0\)
\(\Rightarrow6-2y=0\)
\(\Rightarrow y=3\) ( L ; do y < 3 )
Vậy \(y\ge3\) thì PT vô số nghiệm
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Xét tính đồng biến, nghịch biến:
a) \(y=\dfrac{x^2+2}{x+1}\)
b) \(y=\dfrac{2x^2-3}{x-2}\)
c) \(y=\dfrac{x+1}{x^2-4}\)
d) \(y=\dfrac{2x+3}{x^2-1}\)
Tính:
a) \(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3}\); b) \({y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25)\);
c) \((2{x^2} + x + 4)({x^2} - x - 1)\); d) \((3x - 4)(2x + 1) - (x - 2)(6x + 3)\).
a)
\(\dfrac{1}{2}{x^2}.\dfrac{6}{5}{x^3} = \dfrac{1}{2}.\dfrac{6}{5}.{x^2}.{x^3} = \dfrac{3}{5}{x^5}\);
b)
\(\begin{array}{l}{y^2}(\dfrac{5}{7}{y^3} - 2{y^2} + 0,25) = {y^2}.\dfrac{5}{7}{y^3} - {y^2}.2{y^2} + {y^2}.0,25)\\ = \dfrac{5}{7}{y^5} - 2{y^4} + 0,25{y^2}\end{array}\);
c)
\(\begin{array}{l}(2{x^2} + x + 4)({x^2} - x - 1) \\= 2{x^2}({x^2} - x - 1) + x({x^2} - x - 1) + 4({x^2} - x - 1)\\ = 2{x^4} - 2{x^3} - 2{x^2} + {x^3} - {x^2} - x + 4{x^2} - 4x - 4 \\= 2{x^4} - {x^3} + {x^2} - 5x - 4\end{array}\);
d)
\(\begin{array}{l}(3x - 4)(2x + 1) - (x - 2)(6x + 3) \\= 3x(2x + 1) - 4(2x + 1) - x(6x + 3) + 2(6x + 3)\\ = 6{x^2} + 3x - 8x - 4 - 6{x^2} - 3x + 12x + 6\\ = 4x + 2\end{array}\).
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Chứng minh đẳng thức:
a, \(\left(\dfrac{3}{2x-y}-\dfrac{2}{2x+y}-\dfrac{1}{2x-5y}\right).\dfrac{4x^2-y^2}{y^2}=\dfrac{-24}{2x-5y}\)
b, \(\dfrac{x^2-x+1}{x^2+x}.\dfrac{x+1}{3x-2}.\dfrac{9x-6}{x^2-x+1}=\dfrac{3}{x}\)
Lời giải
a)
\(\left(\frac{3}{2x-y}-\frac{2}{2x+y}-\frac{1}{2x-5y}\right).\frac{4x^2-y^2}{y^2}\)
\(=\frac{3(4x^2-y^2)}{(2x-y)y^2}-\frac{2(4x^2-y^2)}{(2x+y)y^2}-\frac{4x^2-y^2}{(2x-5y)y^2}\)
\(=\frac{3(2x-y)(2x+y)}{(2x-y)y^2}-\frac{2(2x-y)(2x+y)}{(2x+y)y^2}-\frac{4x^2-y^2}{(2x-5y)y^2}\)
\(=\frac{3(2x+y)-2(2x-y)}{y^2}-\frac{4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{2x+5y}{y^2}-\frac{4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{(2x+5y)(2x-5y)-4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}\)
\(=\frac{4x^2-25y^2-4x^2}{(2x-5y)y^2}+\frac{1}{2x-5y}=\frac{-25}{2x-5y}+\frac{1}{2x-5y}=\frac{-24}{2x-5y}\)
Ta có đpcm.
b)
\(\frac{x^2-x+1}{x^2+x}.\frac{x+1}{3x-2}.\frac{9x-6}{x^2-x+1}\)
\(=\frac{(x^2-x+1)(x+1).3(3x-2)}{x(x+1)(3x-2)(x^2-x+1)}\)
\(=\frac{3}{x}\) (đpcm)
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Bài 1 : Tìm x,y,z biết :
a) 2x 3y ; 5y 7z và 3x - 7y + 5z -30
b) 3x 5y ; 7y 2z và x + y + z 74
c) x : z dfrac{2}{3} : dfrac{1}{2} ; z : y 1 : dfrac{4}{7} và y + z 66
d) x : y : z 3 : 4 : 5 và 2x^2 + 2y^2 - 3z^2 -100
e) x:y:z 2 : 5 : 6 và 2x^2 + 4y^2 - 4z^2 -324
f) dfrac{x-1}{2} dfrac{y-2}{3} dfrac{z-3}{4} và x-2y+3z14
g)dfrac{x-1}{2} dfrac{y+3}{4} dfrac{z-5}{6} và 5z-3x-4y50
h) dfrac{x}{2}dfrac{y}{7} và xy56
i)dfrac{x-y}{3}dfrac{x+y}{13}dfrac{xy}{200}
k) dfrac{x-5}{6}d...
Đọc tiếp
Bài 1 : Tìm x,y,z biết :
a) 2x = 3y ; 5y = 7z và 3x - 7y + 5z = -30
b) 3x =5y ; 7y = 2z và x + y + z = 74
c) x : z = \(\dfrac{2}{3}\) : \(\dfrac{1}{2}\) ; z : y = 1 : \(\dfrac{4}{7}\) và y + z = 66
d) x : y : z = 3 : 4 : 5 và \(2x^2\) + \(2y^2\) - \(3z^2\) = -100
e) \(x:y:z\) = 2 : 5 : 6 và \(2x^2\) + \(4y^2\) - \(4z^2\) = -324
f) \(\dfrac{x-1}{2}\) = \(\dfrac{y-2}{3}\) = \(\dfrac{z-3}{4}\) và \(x-2y+3z=14\)
g)\(\dfrac{x-1}{2}\) = \(\dfrac{y+3}{4}\) =\(\dfrac{z-5}{6}\) và \(5z-3x-4y=50\)
h) \(\dfrac{x}{2}=\dfrac{y}{7}\) và \(xy=56\)
i)\(\dfrac{x-y}{3}=\dfrac{x+y}{13}=\dfrac{xy}{200}\)
k) \(\dfrac{x-5}{6}=\dfrac{x+5}{18}\)
l) \(\dfrac{2x-11}{12}=\dfrac{x+5}{20}\)
