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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 \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)
\(\Rightarrow2021^{2022}-1⋮6\)
\(\Rightarrow S-1⋮6\)
Ta có: VT = \(\dfrac{x+1}{2021}\)+1 - (\(\dfrac{x+2}{2020}\)+1) = \(\dfrac{x+3}{2019}\)+1=VP
=>\(\dfrac{x+2022}{2021}+\dfrac{x+2022}{2020}-\dfrac{x+2022}{2019}=0\)
=>\(\left(x+2022\right)\left(\dfrac{1}{2021}+\dfrac{1}{2020}-\dfrac{1}{2019}\right)=0\)
=>x +2022 = 0=> x =-2022
\(\dfrac{1}{x+1}\)-\(\dfrac{5}{x-2}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)\(\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}\)-\(\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}\)=\(\dfrac{15}{\left(x+1\right)\left(x-2\right)}\)
\(\Leftrightarrow\)x-2-5(x+1)=15
\(\Leftrightarrow\) x-2-5x-5=15
\(\Leftrightarrow\)x-5x=15+2+5
\(\Leftrightarrow\)-4x=22
\(\Leftrightarrow\)x=-\(\dfrac{11}{2}\)
vậy
Lời giải :
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=\dfrac{x^2+y^2+z^2}{a^2+b^2+c^2}\Leftrightarrow\dfrac{x^2}{a^2+b^2+c^2}-\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2+b^2+c^2}-\dfrac{y^2}{b^2}+\dfrac{z^2}{a^2+b^2+c^2}-\dfrac{z^2}{c^2}=0\)
\(\Leftrightarrow x^2\left(\dfrac{1}{a^2+b^2+c^2}-\dfrac{1}{a^2}\right)+y^2\left(\dfrac{1}{a^2+b^2+c^2}-\dfrac{1}{b^2}\right)+z^2\left(\dfrac{1}{a^2+b^2+c^2}-\dfrac{1}{c^2}\right)=0\)
Do \(\dfrac{1}{a^2+b^2+c^2}-\dfrac{1}{a^2}\ne0;\dfrac{1}{a^2+b^2+c^2}-\dfrac{1}{b^2}\ne0;\dfrac{1}{a^2+b^2+c^2}-\dfrac{1}{c^2}\ne0\)
\(\Rightarrow\) \(\left\{{}\begin{matrix}x^2=0\\y^2=0\\z^2=0\end{matrix}\right.\) \(\Rightarrow\)\(\left\{{}\begin{matrix}x=0\\y=0\\z=0\end{matrix}\right.\)
Thay vào biểu thức P :
\(P=0^{2020}+\left(y-1\right)^{2022}+\left(z-1\right)^{203}=0+1-1=0\)
1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(3x+9+4x-12=3x-7\)
\(\Leftrightarrow4x=-7+12-9=-4\)
hay \(x=-1\left(nhận\right)\)
2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(3x+12-4x+16=3x-4\)
\(\Leftrightarrow28-4x=-4\)
\(\Leftrightarrow4x=32\)
hay \(x=8\left(tm\right)\)
3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
Suy ra: \(5x^2-12+3x+3=5x^2-5x\)
\(\Leftrightarrow3x-9+5x=0\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(nhận\right)\)
1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)
\(\Leftrightarrow-3x-12-3+5x-x+4=0\)
\(\Leftrightarrow x=11\left(nhận\right)\)
2. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{3(x-2)-(x-1)}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)
\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)
Vậy pt vô nghiệm
\(\dfrac{x-1}{2023}+\dfrac{x-2}{2022}=\dfrac{x-3}{2021}+\dfrac{x-4}{2020}\)
`<=>(x-1)/2023-1+(x-2)/2022-1=(x-3)/2021-1+(x-4)/2020-1`
`<=>(x-2024)/2023+(x-2024)/2022=(x-2024)/2021+(x-2024)/2020`
`<=>(x-2024)(1/2023+1/2022-1/2021-1/2020)=0`
`<=>x-2024=0(1/2023+1/2022-1/2021-1/2020>0)`
`<=>x=2024`
=>\(\left(\dfrac{x-1}{2023}-1\right)+\left(\dfrac{x-2}{2022}-1\right)=\left(\dfrac{x-3}{2021}-1\right)+\left(\dfrac{x-4}{2020}-1\right)\)
=>x-2024=0
=>x=2024