Cho biểu thức M=(\(\dfrac{2x}{x+3}\dfrac{x}{x-3}-\dfrac{3x^2+3}{x^2-9}\))\(\div\)(\(\dfrac{2x-2}{x-3}-1\)) ( x\(\ne\)\(\pm\)3 ; x\(\ne\)-1)
1. Rút gọn M
2. Tìm X để M\(\le\)\(\dfrac{-1}{2}\)
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: ...
\(\Leftrightarrow\dfrac{3x+2022+2x-2021}{\left(2x-2021\right)\left(3x+2022\right)}=\dfrac{10x-2024-\left(15x-2023\right)}{\left(15x-2023\right)\left(10x-2024\right)}\)
\(\Leftrightarrow\dfrac{5x-1}{\left(2x-2021\right)\left(3x+2022\right)}=-\dfrac{5x-1}{\left(15x-2023\right)\left(10x-2024\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\Rightarrow x=...\\\dfrac{1}{\left(2x-2021\right)\left(3x+2022\right)}=-\dfrac{1}{\left(15x-2023\right)\left(10x-2024\right)}\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\left(2x-2021\right)\left(3x+2022\right)+\left(15x-2023\right)\left(10x-2024\right)=0\)
\(\Leftrightarrow\left[12x-4045-\left(10x-2024\right)\right]\left(3x+2022\right)+\left(12x-4045+3x+2022\right)\left(10x-2024\right)=0\)
\(\Leftrightarrow\left(12x-4045\right)\left(3x+2022\right)-\left(10x-2024\right)\left(3x+2022\right)+\left(12x-4045\right)\left(10x-2024\right)+\left(3x+2022\right)\left(10x-2024\right)=0\)
\(\Leftrightarrow\left(12x-4045\right)\left(13x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{13}\\x=\dfrac{4045}{12}\end{matrix}\right.\)
Thực hiện phép tính:
a) \(\dfrac{x+2y}{xy}\div\dfrac{x^2+4xy+4y^2}{2x^2}\)
b) \(\dfrac{4x^3-xy^2}{x^2+xy+y^2}\div\dfrac{\left(2x-y\right)^3}{x^3-y^3}\)
c) \(\dfrac{x+3}{x+2}\div\dfrac{3x+9}{2x-1}\div\dfrac{4x-2}{2x+4}\)
d) \(\dfrac{x+1}{x+2}\div\left(\dfrac{2x^2}{2x-3}\times\dfrac{3x+3}{4x^3}\right)\)
a: \(=\dfrac{x+2y}{xy}\cdot\dfrac{2x^2}{\left(x+2y\right)^2}=\dfrac{2x}{y\left(x+2y\right)}\)
b: \(=\dfrac{x\left(4x^2-y^2\right)}{x^2+xy+y^2}\cdot\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)\left(2x-y\right)}{\left(2x-y\right)^3}\)
\(=\dfrac{x\left(x-y\right)\left(2x+y\right)}{\left(2x-y\right)^2}\)
c: \(=\dfrac{x+3}{x+2}\cdot\dfrac{2x-1}{3\left(x+3\right)}\cdot\dfrac{2\left(x+2\right)}{2\left(2x-1\right)}\)
=1/3
d: \(=\dfrac{x+1}{x+2}:\left(\dfrac{1}{2x}\cdot\dfrac{3x+3}{2x-3}\right)\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{2x\left(2x-3\right)}{3\left(x+1\right)}=\dfrac{2x\left(2x-3\right)}{3\left(x+2\right)}\)
Cho biểu thức A=\(\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\)
và B=\(\dfrac{x^2+x-2}{x^3-1}\)
a Rút gọn biểu thức M=A.B
b Tìm x thuộc Z để M thuộc Z
c Tìm GTLN của biểu thức N=\(A^{-1}-B\)
a. \(A=\left(\dfrac{2-3x}{x^2+2x-3}-\dfrac{x+3}{1-x}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{x^3-1}\left(ĐKXĐ:x\ne1;x\ne-3\right)\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{x-1}-\dfrac{x+1}{x+3}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\left(\dfrac{2-3x}{\left(x-1\right)\left(x+3\right)}+\dfrac{\left(x+3\right)^2}{\left(x-1\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+3\right)}\right):\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{2-3x+x^2+6x+9-x^2+1}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}:\dfrac{3x+12}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{3x+12}{\left(x-1\right)\left(x+3\right)}.\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{3x+12}=\dfrac{x^2+x+1}{x+3}\)
\(M=A.B=\dfrac{x^2+x+1}{x+3}.\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x^2+x-2}{x+3}\)
b. -Để M thuộc Z thì:
\(\left(x^2+x-2\right)⋮\left(x+3\right)\)
\(\Rightarrow\left(x^2+3x-2x-6+4\right)⋮\left(x+3\right)\)
\(\Rightarrow\left[x\left(x+3\right)-2\left(x+3\right)+4\right]⋮\left(x+3\right)\)
\(\Rightarrow4⋮\left(x+3\right)\)
\(\Rightarrow x+3\in\left\{1;2;4;-1;-2;-4\right\}\)
\(\Rightarrow x\in\left\{-2;-1;1;-4;-5;-7\right\}\)
c. \(A^{-1}-B=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{x^3-1}\)
\(=\dfrac{x+3}{x^2+x+1}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{x^2+x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^2-x+3x-3-x^2-x+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
\(=\dfrac{1}{x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\)
\(Max=\dfrac{4}{3}\Leftrightarrow x=\dfrac{-1}{2}\)
1) Chứng minh biểu thứ A = ( \(\dfrac{2x^3+2}{x+1}-2x\))(\(\dfrac{x^3-1}{x-1}+x\)) ( x \(\ne\)1 và -2) luôn luôn dương với mọi x\(\ne\)\(\pm\)1
2) Tìm Min của biểu thức y = \(\dfrac{x^4+4x^2+10}{x^4+6x^2+9}\)
1 ) \(A=\left(\dfrac{2x^3+2}{x+1}-2x\right)\left(\dfrac{x^3-1}{x-1}+x\right)\)
\(\Leftrightarrow A=\left(\dfrac{2x^3+2-2x^2-2x}{x+1}\right)\left(x^2+2x+1\right)\)
\(\Leftrightarrow A=\left(\dfrac{\left(2x^2-2\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=\left(\dfrac{2\left(x-1\right)\left(x+1\right)\left(x-1\right)}{x+1}\right)\left(x+1\right)^2\)
\(\Leftrightarrow A=2\left(x-1\right)^2\left(x+1\right)^2\ge0\forall x\)
BÀI 1 : RÚT GỌN CÁC BIỂU THỨC SAU .
a, \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}\)
b, \(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{9x^2-6x+1}\)
c, \(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
d, \(\dfrac{1-x^2}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
Câu 10 (2,5 điểm). Cho hai biểu thức $P=\dfrac{2{{x}^{2}}-1}{{{x}^{2}}+x}-\dfrac{x-1}{x}+\dfrac{3}{x+1}$ với $x\ne 0, \, x\ne -1$ và $Q=\dfrac{x+1}{{{x}^{2}}-9}$ với $x\ne \pm 3$.
a) Tính giá trị biểu thức $Q$ khi $x=2$.
b) Rút gọn biểu thức $P$.
c) Đặt $M=P.Q$. Tìm $x$ để $M=\dfrac{-1}{2}$.
a)Thay x=2(TMDK) vào bt Q :
\(Q=\dfrac{2+1}{2^2-9}=-\dfrac{3}{5}\)
b) \(P=\dfrac{2x^2-1}{x^2+x}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\\ =\dfrac{2x^2-1}{x\left(x+1\right)}-\dfrac{x-1}{x}+\dfrac{3}{x+1}\\ =\dfrac{2x^2-1-\left(x-1\right)\left(x+1\right)+3x}{x\left(x+1\right)}\\ =\dfrac{2x^2-1-\left(x^2-1\right)+3x}{x\left(x+1\right)}\\ =\dfrac{x^2+3x}{x\left(x+1\right)}=\dfrac{x\left(x+3\right)}{x\left(x+1\right)}=\dfrac{x+3}{x+1}\)
c) \(M=P.Q=\dfrac{x+3}{x+1}.\dfrac{x+1}{x^2-9}\\ =\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{1}{x-3}\)
\(M=-\dfrac{1}{2}\\ =>\dfrac{1}{x-3}=-\dfrac{1}{2}\\ =>x-3=-2\\ =>x=1\left(TMDK\right)\)
Chứng minh rằng :
a) Giá trị của biểu thức :
\(\left(\dfrac{x+1}{x}\right)^2:\left[\dfrac{x^2+1}{x^2}+\dfrac{2}{x+1}\left(\dfrac{1}{x}+1\right)\right]\) bằng 1 với mọi giá trị \(x\ne0;x\ne-1\)
b) Giá trị của biểu thức :
\(\dfrac{x}{x-3}-\dfrac{x^2+3x}{2x+3}\left(\dfrac{x+3}{x^2-3x}-\dfrac{x}{x^2-9}\right)\) bằng 1 khi \(x\ne0;x\ne-3;x\ne3;x\ne-\dfrac{3}{2}\)
Cho các biểu thức : A= \(\left(-\dfrac{3x+15}{x^2+10x+25}\right):\left(\dfrac{x}{x+3}-\dfrac{2x}{3-x}-\dfrac{3x^2+9}{x^2-9}\right)
\)
a). Tìm điều kiện xác định và rút gọn biểu thức A
b). Tìm x để A<1
c). Tìm x để A = \(\dfrac{2x-3}{x+1}\)
a: ĐKXĐ: \(x\in\left\{-5;3;-3\right\}\)
\(A=\dfrac{-3\left(x+5\right)}{\left(x+5\right)^2}:\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{-3}{x+5}\cdot\dfrac{\left(x-3\right)\left(x+3\right)}{-3\left(x+3\right)}\)
\(=\dfrac{x-3}{x+5}\)
b: Để A<1 thì A-1<0
=>\(\dfrac{x-3-x-5}{x+5}< 0\)
=>x+5>0
=>x>-5
c: Để A=(2x-3)/(x+1) thì \(\dfrac{2x-3}{x+1}=\dfrac{x-3}{x+5}\)
=>2x^2+10x-3x-15=x^2-2x-3
=>2x^2+7x-15-x^2+2x+3=0
=>x^2+9x-12=0
hay \(x=\dfrac{-9\pm\sqrt{129}}{2}\)
thực hiện phép tính
\(\dfrac{4x^2-3x+5}{x^3-1}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)
\(\dfrac{15x-11}{x^2+2x-3}-\dfrac{3x-2}{x-1}-\dfrac{2x+3}{3+x}\)
\(\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{2x\left(1-x\right)}{9-x^2}\)
\(\dfrac{4x^2-3x+5}{x^3-1}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)
\(\Leftrightarrow\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1+2x}{x^2+x+1}-\dfrac{6}{x-1}\)
\(ĐKXĐ:x\ne1\)
\(\dfrac{4x^2-3x+5}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{(1+2x)\left(x-1\right)}{(x^2+x+1)\left(x-1\right)}-\dfrac{6\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\)
\(\Rightarrow4x^2-3x+5-\left(1+2x\right)\left(x-1\right)-6\left(x^2+x+1\right)\)
\(\Rightarrow4x^2-3x+5-\left(x-1+2x^2-2x\right)-6x^2-6x-6\)
\(\Rightarrow4x^2-3x+5-x+1-2x^2+2x-6x^2-6x-6\)
\(\Rightarrow-4x^2-8x\)
⇒-4x(x-4)