\(x^2+1-\dfrac{x^4+1}{x^2+1}=\dfrac{\left(x^2+1\right)^2-x^4-1}{x^2+1}=\dfrac{x^4+2x^2+1-x^4-1}{x^2+1}=\dfrac{2x^2}{x^2+1}\)
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B1: Aleft(dfrac{2-3x}{x^2+2x-3}-dfrac{x+3}{1-x}-dfrac{x+1}{x+3}right):dfrac{3x+12}{x^3-1}a) Rút gọnb) Tìm x thuộc Z để A nguyênc) Tính A với x-2; x-3d) Tìm x dể A1B2: Phân tích thành nhân tửa) x2-2xy-4+y2b) x2-4x+3c) 9x2(x-y)-x+yB3: Rút gọna) (x-2)3-(x+2)3-(x-1)(x2+x+1)b) (5x+3y)(5x-3y)+(4x-3y)2B4: P(x)x4+x3+mx2-3x+5a) Khi m4, thực hiện phép chia P(x) cho x2-x+1b) Tìm m để P(x)⋮(x-1)
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B1: A=\(\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\)
a) Rút gọn
b) Tìm x thuộc Z để A nguyên
c) Tính A với x=-2; x=-3
d) Tìm x dể A=1
B2: Phân tích thành nhân tử
a) x2-2xy-4+y2
b) x2-4x+3
c) 9x2(x-y)-x+y
B3: Rút gọn
a) (x-2)3-(x+2)3-(x-1)(x2+x+1)
b) (5x+3y)(5x-3y)+(4x-3y)2
B4: P(x)=x4+x3+mx2-3x+5
a) Khi m=4, thực hiện phép chia P(x) cho x2-x+1
b) Tìm m để P(x)⋮(x-1)
dùng định nghĩa hai phân thức bằng nhau chứng tỏ rằng :
a,dfrac{x2y2}{5}dfrac{7x3y4}{35xy}
b,dfrac{x3-4x}{10-5x}dfrac{-X2-2X}{5}
C,dfrac{x+2}{X-1}dfrac{left(x+2right)left(x+1right)}{x2-1}
d,dfrac{x2-x-2}{x+1}dfrac{x2-3x+2}{x-1}
e,dfrac{x3+8}{x2-2x+4}x+2
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dùng định nghĩa hai phân thức bằng nhau chứng tỏ rằng :
a,\(\dfrac{x2y2}{5}\)=\(\dfrac{7x3y4}{35xy}\)
b,\(\dfrac{x3-4x}{10-5x}\)=\(\dfrac{-X2-2X}{5}\)
C,\(\dfrac{x+2}{X-1}\)=\(\dfrac{\left(x+2\right)\left(x+1\right)}{x2-1}\)
d,\(\dfrac{x2-x-2}{x+1}\)=\(\dfrac{x2-3x+2}{x-1}\)
e,\(\dfrac{x3+8}{x2-2x+4}\)=x+2
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 phương trình sau :
a, \(\dfrac{1}{x+1}+\dfrac{2}{x^3-x^2-x+1}+\dfrac{3}{x^2-1}=0\)
b, \(\dfrac{1}{x^2+5x+4}+\dfrac{1}{x^2+11x+28}+\dfrac{1}{x^2+17x+70}+\dfrac{1}{x^2+23x+130}=\dfrac{4}{13}\)
\(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)
4.Giải phương trình
a) dfrac{x+5}{x-5}-dfrac{x-5}{x+5}dfrac{20}{x^2-25}
b)dfrac{1}{x-1}+dfrac{2}{x+1}dfrac{x}{x^2-1}
c)5+dfrac{76}{x^2-16}dfrac{2x-1}{x+4}-dfrac{3x-1}{4-x}
d)dfrac{90}{x}-dfrac{36}{x-6}2
e)dfrac{1}{x}+dfrac{1}{x+10}dfrac{1}{12}
f)dfrac{x+3}{x-3}-dfrac{1}{x}dfrac{3}{xleft(x-3right)}
g)dfrac{3}{x+2}-dfrac{2}{x-2}+dfrac{8}{x^2-4}0
h)dfrac{3}{x+2}-dfrac{2}{x-3}dfrac{8}{left(x-3right)left(x+2right)}
i)dfrac{x}{2x+6}-dfrac{x}{2x+2}dfrac{3x+2}{left(x+1right)left(x+3right)}
k)d...
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4.Giải phương trình
a) \(\dfrac{x+5}{x-5}-\dfrac{x-5}{x+5}=\dfrac{20}{x^2-25}\)
b)\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{x}{x^2-1}\)
c)\(5+\dfrac{76}{x^2-16}=\dfrac{2x-1}{x+4}-\dfrac{3x-1}{4-x}\)
d)\(\dfrac{90}{x}-\dfrac{36}{x-6}=2\)
e)\(\dfrac{1}{x}+\dfrac{1}{x+10}=\dfrac{1}{12}\)
f)\(\dfrac{x+3}{x-3}-\dfrac{1}{x}=\dfrac{3}{x\left(x-3\right)}\)
g)\(\dfrac{3}{x+2}-\dfrac{2}{x-2}+\dfrac{8}{x^2-4}=0\)
h)\(\dfrac{3}{x+2}-\dfrac{2}{x-3}=\dfrac{8}{\left(x-3\right)\left(x+2\right)}\)
i)\(\dfrac{x}{2x+6}-\dfrac{x}{2x+2}=\dfrac{3x+2}{\left(x+1\right)\left(x+3\right)}\)
k)\(\dfrac{x}{x+1}-\dfrac{2x-3}{1-x}=\dfrac{3x^2+5}{x^2-1}\)
l)\(\dfrac{5}{x+7}+\dfrac{8}{2x+14}=\dfrac{3}{2}\)
m)\(\dfrac{x-1}{x}-\dfrac{1}{x+1}=\dfrac{2x-1}{x^2+x}\)
Cần gấp ạ
Giải phương trình:\(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x+\dfrac{1}{x}\right)^2\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\dfrac{1}{\sqrt{x}+\sqrt{x+2}}+\dfrac{1}{\sqrt{x+2}+\sqrt{x+4}}+\dfrac{1}{\sqrt{x+4}+\sqrt{x+6}}=\dfrac{\sqrt{10}}{2}-1\)
ta có Adfrac{1}{1+dfrac{bc}{a}}+dfrac{1}{1+dfrac{ca}{b}}+dfrac{1}{1+dfrac{ab}{c}}
đặt sqrt{dfrac{bc}{a}};sqrt{dfrac{ca}{b}};sqrt{dfrac{ab}{c}}left(x;y;zright) xy+yz+zx1
ta có Adfrac{1}{1+x^2}+dfrac{1}{1+y^2}+dfrac{1}{1+z^2}
ta cần chứng minh dfrac{1}{1+x^2}+dfrac{1}{1+y^2}+dfrac{1}{1+z^2}gedfrac{9}{4}Leftrightarrow1-dfrac{1}{x^2}+1-dfrac{1}{1+y^2}+1-dfrac{1}{z^2+1}ledfrac{3}{4}
Leftrightarrowdfrac{x^2}{x^2+1}+dfrac{y^2}{y^2+1}+dfrac{z^2}{z^2+1}gedfrac{3}{4}
mà dfrac{x^2}{x^2+1}+dfrac{y^2}{...
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ta có \(A=\dfrac{1}{1+\dfrac{bc}{a}}+\dfrac{1}{1+\dfrac{ca}{b}}+\dfrac{1}{1+\dfrac{ab}{c}}\)
đặt \(\sqrt{\dfrac{bc}{a}};\sqrt{\dfrac{ca}{b}};\sqrt{\dfrac{ab}{c}}=\left(x;y;z\right)\) =>xy+yz+zx=1
ta có A=\(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{1}{1+z^2}\)
ta cần chứng minh \(\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{1}{1+z^2}\ge\dfrac{9}{4}\Leftrightarrow1-\dfrac{1}{x^2}+1-\dfrac{1}{1+y^2}+1-\dfrac{1}{z^2+1}\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{x^2}{x^2+1}+\dfrac{y^2}{y^2+1}+\dfrac{z^2}{z^2+1}\ge\dfrac{3}{4}\)
mà \(\dfrac{x^2}{x^2+1}+\dfrac{y^2}{y^2+1}+\dfrac{z^2}{z^2+1}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+3}=\dfrac{x^2+y^2+z^2+2}{x^2+y^2+z^2+3}=1-\dfrac{1}{x^2+y^2+z^2+3}\ge\dfrac{3}{4}\)
=> BĐT cầnd chứng minh luôn đúng
Tìm x
a)(2x+1)2-4(x+2)2 =9
b)(3x-1)2 +2(x+3)2 +11(x+1)(1-x)=6
c)(x+1)3 - x2 (x+3)=2
d)(x-2)3 -x(x+1)(x-1)+6x2 =5
e)(x-3)(x2 +3x +9)-x(x+4)(x-4)=5
g)(x-2)3 -(x+5)(x2 -5x+25)+6x2 =11