Rút gọn biểu thức :
a) \(\dfrac{x^4+15x+7}{2x^3+2}.\dfrac{x}{14x^2+1}.\dfrac{4x^3+4}{x^4+15x+7}\)
b) \(\dfrac{x^7+3x^2+2}{x^3-1}.\dfrac{3x}{x+1}.\dfrac{x^2+x+1}{x^7+3x^2+2}\)
Câu 1 :
Cho biểu thức \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right).\dfrac{7}{x^2+8}vớix\ne\pm\sqrt{3}\)
1.Rút gọn P
2.Tìm x để P nhận giá trị nguyên
Câu 2 :
1.Giải phương trình : \(\dfrac{1}{2x-2021}+\dfrac{1}{3x+2022}=\dfrac{1}{15x-2023}-\dfrac{1}{10x-2024}\)
2.Cho đa thức \(P\left(x\right)=2x^3-x^2+ax+bvàQ\left(x\right)=x^2-4x+4\).Tìm a,b để đa thức P(x) chia hết cho đa thức Q(x)
Câu 3:
1.Cho hai số thực x,y thỏa mãn \(0< xy\le1\) . Chứng minh \(\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}\le\dfrac{2}{xy+1}\)
2.Cho \(S=a^3_1+a^3_2+a^3_3+...+a^3_{100}\) với \(a_1,a_2,a_3,...a_{100}\) là các số nguyên thỏa mãn \(a_1+a_2+a_3+...+a_{100}=2021^{2022}.CMR:S-1⋮6\)
Câu 1:
1: Ta có: \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\left(\dfrac{x^2\left(x^2+3\right)}{\left(x^2-3\right)\left(x^2+3\right)}+\dfrac{2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\right)\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+3x^2+2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+5x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^4+8x^2-3x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{x^2\left(x^2+8\right)-3\left(x^2+8\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{\left(x^2+8\right)\left(x^2-3\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)
\(=\dfrac{7}{x^2+3}\)
Câu 2a đề sai, pt này ko giải được
2b.
\(P\left(x\right)=\left(2x+7\right)\left(x^2-4x+4\right)+\left(a+20\right)x+\left(b-28\right)\)
Do \(\left(2x+7\right)\left(x^2-4x+4\right)⋮\left(x^2-4x+4\right)\)
\(\Rightarrow P\left(x\right)\) chia hết \(Q\left(x\right)\) khi \(\left(a+20\right)x+\left(b-28\right)\) chia hết \(x^2-4x+4\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+20=0\\b-28=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-20\\b=28\end{matrix}\right.\)
3a.
\(VT=\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}=1+\dfrac{1-x^2y^2}{1+x^2+y^2+x^2y^2}\le1+\dfrac{1-x^2y^2}{1+2xy+x^2y^2}\)
\(VT\le1+\dfrac{\left(1-xy\right)\left(1+xy\right)}{\left(xy+1\right)^2}=1+\dfrac{1-xy}{1+xy}=\dfrac{2}{1+xy}\) (đpcm)
3b
Ta có: \(n^3-n=n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6
\(\Rightarrow n^3\) luôn đồng dư với n khi chia 6
\(\Rightarrow S\equiv2021^{2022}\left(mod6\right)\)
Mà \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)
\(\Rightarrow2021^{2022}-1⋮6\)
\(\Rightarrow S-1⋮6\)
2a.
À nãy mình nhìn lộn dấu trừ bên vế phải thành dấu cộng
ĐKXĐ: ...
\(\Leftrightarrow\dfrac{3x+2022+2x-2021}{\left(2x-2021\right)\left(3x+2022\right)}=\dfrac{10x-2024-\left(15x-2023\right)}{\left(15x-2023\right)\left(10x-2024\right)}\)
\(\Leftrightarrow\dfrac{5x-1}{\left(2x-2021\right)\left(3x+2022\right)}=-\dfrac{5x-1}{\left(15x-2023\right)\left(10x-2024\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\Rightarrow x=...\\\dfrac{1}{\left(2x-2021\right)\left(3x+2022\right)}=-\dfrac{1}{\left(15x-2023\right)\left(10x-2024\right)}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(2x-2021\right)\left(3x+2022\right)+\left(15x-2023\right)\left(10x-2024\right)=0\)
\(\Leftrightarrow\left[12x-4045-\left(10x-2024\right)\right]\left(3x+2022\right)+\left(12x-4045+3x+2022\right)\left(10x-2024\right)=0\)
\(\Leftrightarrow\left(12x-4045\right)\left(3x+2022\right)-\left(10x-2024\right)\left(3x+2022\right)+\left(12x-4045\right)\left(10x-2024\right)+\left(3x+2022\right)\left(10x-2024\right)=0\)
\(\Leftrightarrow\left(12x-4045\right)\left(13x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{13}\\x=\dfrac{4045}{12}\end{matrix}\right.\)
BÀI 6 :rút gọn phân thức
\(\dfrac{x^3+3x^3+3x+1}{x^2+x}\)
b)\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)
c)\(\dfrac{x^2+4x+4}{2x+4}\)
d)\(\dfrac{(x-1)(-x-2)}{x+2}\)
e)\(\dfrac{x^2-y^2}{x+y}\)
f)\(\dfrac{3x^2+4xy^2}{6x+8y}\)
g)\(\dfrac{-3x^2-6x}{4-x^2}\)
BÀI 7 :quy đồng mẫu thức các phân thức
\(\dfrac{2}{5x^3y^2}và \dfrac{3}{4xy}\)
b)\(\dfrac{x}{x^2-2xy+y^2} và \dfrac{x}{x^2-xy}\)
c)\(\dfrac{1}{x+2};\dfrac{2}{2x+4}và \dfrac{3}{3x+6}\)
d)\(\dfrac{1}{x+3};\dfrac{2}{2x-6}và \dfrac{3}{3x-9}\)
6:
a: ĐKXĐ: x<>0
\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)
\(=\dfrac{\left(x+1\right)^3}{x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x}\)
b: ĐKXĐ: x<>1
\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)
\(=\dfrac{\left(x-1\right)^3}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{2}\)
c: ĐKXĐ: x<>-2
\(\dfrac{x^2+4x+4}{2x+4}\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}\)
\(=\dfrac{x+2}{2}\)
d: ĐKXĐ: x<>-2
\(\dfrac{\left(x-1\right)\left(-x-2\right)}{x+2}\)
\(=\dfrac{\left(-x+1\right)\left(x+2\right)}{x+2}=-x+1\)
e: ĐKXĐ: x<>-y
\(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)
g: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)
\(=\dfrac{3x\left(x+2\right)}{\left(x+2\right)\cdot\left(x-2\right)}=\dfrac{3x}{x-2}\)
7:
a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)
\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)
b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)
\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)
c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)
\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)
d:
\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)
\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)
\(\dfrac{1}{x+3}=\dfrac{x-3}{\left(x+3\right)\left(x-3\right)}\)
Thực hiện phép tính :
a/ (x - 1)^2 - (4x + 3) (2 - x)
b/ (15x^3y^2 - 6x^2y^3) : 3x^2y^2 = (15x^3y^2 : 3x^2y^2) - (6x^2y^3 : 3x^2y^2) = 5x - 2y
c/\(\dfrac{x+7}{x-7}\) - \(\dfrac{x-7}{x+7}\) +\(\dfrac{4x^2}{x^2-49}\)
a/ (x-1)2-(4x+3)(2-x)=x2-2x+1-(8x-4x2+6-3x)
=x2-2x+1-8x+4x2-6+3x=5x2-7x-6
b/ (15x3y2 - 6x2y3) : 3x2y2 = 5x - 2y
c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)
rút gọn các phân thức
a,\(\dfrac{7xy^3\left(x-2y\right)}{14x^2y^2\left(x-2y\right)^2}\)
b,\(\dfrac{4a^2-8ab}{2\left(2b-a\right)^3}\)
c,\(\dfrac{3x^3-3x}{x^4-1}\)
d,\(\dfrac{45x\left(3-x\right)}{15x\left(x-3\right)^3}\)
c: \(=\dfrac{3x\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}=\dfrac{3x}{x^2+1}\)
Rút gọn biểu thức:
a, \(\frac{x^4+15x+7}{2x^3+2}.\frac{x}{14x^2+1}.\frac{4x^3+4}{x^4+15x+7}\)
b, \(\frac{x^7+3x^2+2}{x^3-1}.\frac{3x}{x+1}.\frac{x^2+x+1}{x^7+3x^2+2}\)
Rút gọn
a) \(\dfrac{x^5-2x^4+2x^3-4x^2-3x+6}{x+4}\)
b) \(\dfrac{x^4-4x^2+3}{x^4+6x^2-7}\)
c) \(\dfrac{x^4+x^3-x-1}{x^4+x^3+2x^2+x+1}\)
\(a,=\dfrac{x^4\left(x-2\right)+2x^2\left(x-2\right)-3\left(x-2\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4+2x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x^4-x^2+3x^2-3\right)}{x+4}\\ =\dfrac{\left(x-2\right)\left(x-1\right)\left(x^2+3\right)}{x+4}\)
\(b,=\dfrac{x^4-3x^2-x^2+3}{x^4-x^2+7x^2-7}=\dfrac{\left(x^2-3\right)\left(x^2-1\right)}{\left(x^2+7\right)\left(x^2-1\right)}=\dfrac{x^2-3}{x^2+7}\\ c,=\dfrac{\left(x^3-1\right)\left(x+1\right)}{x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)}\\ =\dfrac{\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)}{\left(x^2+1\right)\left(x^2+x+1\right)}=\dfrac{x^2-1}{x^2+1}\)
a)\(\dfrac{3x}{2}+\dfrac{4x-3}{10}-\dfrac{15x-1}{5}\)
b)\(\dfrac{x-4}{x+1}+\dfrac{x+1}{x}=2\)
a: \(=\dfrac{5\cdot3x+4x-3-2\left(15x-1\right)}{10}=\dfrac{19x-3-30x+2}{10}=\dfrac{-11x-1}{10}\)
b: \(\Leftrightarrow x\left(x-4\right)+\left(x+1\right)^2=2x\left(x+1\right)\)
\(\Leftrightarrow x^2-4x+x^2+2x+1=2x^2+2x\)
=>-2x+1=2x
=>-4x=-1
hay x=1/4(nhận)
\(a,\dfrac{3x}{2}+\dfrac{4x-3}{10}-\dfrac{15x-1}{5}\)
\(=\dfrac{3x.5+4x-3-2\left(15x-1\right)}{10}\)
\(=\dfrac{15x+4x-3-30x+3}{10}\)
\(=\dfrac{-11x}{10}\)
1.rút gọn biểu thuc P=\(\dfrac{2}{x+3}+\dfrac{1}{x-3}+\dfrac{9-x}{9-x^2}\) với x\(\ne-3vàx\ne3\)
2.thực hiện phép tính \(\left(2x^4-3x^3-3x^2+6x-1\right):\left(x^2-2\right)\)
\(\left(15x^4y^6-12^3y^4-18x^2y^3\right):\left(-6x^2y^2\right)\)
Giải các phương trình:
1/ \(6x^4-x^3-7x^2+x+1=0\)
2/ \(\dfrac{2x-1}{3x^2+7x+2}+\dfrac{3}{9x^2+15x+4}-\dfrac{2x+7}{3x^2-5x-12}=\dfrac{5}{x+2}\)
1/
Ta có: 6x4 -x3-7x2+x+1=0
<=> 6x4-6x3+5x3-5x2-2x2+2x-x+1=0
<=> 6x3(x-1)+5x2(x-1)-2x(x-1)-(x-1)=0
<=> (x-1) ( 6x3+5x2-2x-1)=0
<=> ( x-1) ( 6x3-3x2+8x2-4x+2x-1)=0
<=> (x-1)\(\left[3x^2\left(2x-1\right)+4x\left(2x-1\right)+\left(2x-1\right)\right]\)=0
<=> (x-1) ( 2x-1) ( 3x2+4x+1)=0
<=> (x-1) ( 2x-1) (3x2+3x+x+1)=0
<=> (x-1) (2x-1) \(\left[3x\left(x+1\right)+\left(x+1\right)\right]\)=0
<=> (x-1)(2x-1)(x+1)(3x+1)=0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-1=0\\x+1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=1\\x=-1\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\\x=-1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
vậy \(S=\left\{\pm1;\dfrac{1}{2};\dfrac{-1}{3}\right\}\)
\(6x^4-x^3-7x^2+x+1=0\)
\(\Leftrightarrow6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1=0\)
\(\Leftrightarrow6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x^3+5x^2-2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(6x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=\dfrac{1}{2}\\x=-\dfrac{1}{3}\end{matrix}\right.\)