Cho \(x=\frac{9m^2-4n^2-p^2}{8np}\)
\(y=\frac{\left(2n-p+3m\right)\left(2n-p-3m\right)}{3\left(4n^2+p^2-9m^2+4np\right)}\)
Tính \(Q=\left(6xy+1-2x-3y\right)^5\)
\(hpt:\hept{\begin{cases}3x+2y=-8\\-3x+\left(m+5\right)y=\left(m-1\right)\left(m+1\right)\end{cases}}\)
từ pt 1 \(\Rightarrow y=\frac{-8-3x}{2}\)(3)
thay (3) vào pt 2 ta được
\(-3mx+\left(m+5\right)\left(\frac{-8-3x}{2}\right)=\left(m-1\right)\left(m+1\right)\)
\(\Leftrightarrow-6mx-8m-40-15x-3mx=2\left(m^2-1\right)\)
\(\Leftrightarrow-9mx-15x=2m^2-2+40+8m\)
\(\Leftrightarrow x\left(-9m-15\right)=2m^2+8m+38\)(*)
để hệ phương trình có No duy nhất thì -9m-15\(\ne\)0 \(\Leftrightarrow m\ne\frac{-15}{9}\)
khi đó pt * có No: \(x=-\frac{2m^2+8m+38}{9m+15}\)
với \(x=-\frac{2m^2+8m+38}{9m+15}\)thì \(y=\left(-8+\frac{3\left(2m^2+8m+38\right)}{9m+15}\right):2=\frac{-8\left(9m+15\right)+3\left(2m^2+8m+38\right)}{9m+15}.\frac{1}{2}\)
\(=\frac{-72m-120+6m^2+24m+114}{9m+15}.\frac{1}{2}=\frac{6m^2-48m-6}{9m+15}.\frac{1}{2}=\frac{2\left(3m^2-24m-3\right)}{9m+15}.\frac{1}{2}=\frac{3m^2-24m-3}{9m+15}\)
lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)
lim \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\)
lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
a) lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)
= lim \(\frac{\left(2-\frac{3}{n}+\frac{5}{n^2}\right)\left(2+\frac{1}{n}\right)}{\left(\frac{4}{n}-3\right)\left(2+\frac{1}{n}+\frac{1}{n^2}\right)}=\frac{4}{-6}=-\frac{2}{3}\)
b)lim ( \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\))
= lim ( \(\frac{n\sqrt{n^4+1}-\sqrt{4n^6+2}}{n^2}\) )
= lim \(\frac{\left(n^6+n^2\right)-\left(4n^6+2\right)}{n^2\left(n\sqrt{n^4+1}+\sqrt{4n^2+2}\right)}\)
= lim \(\frac{-3n^6+n^2+2}{n^3\sqrt{n^4+1}+n^2\sqrt{4n^2+2}}\)
= lim \(\frac{-3n\left(1-\frac{1}{n^4}-\frac{2}{n^6}\right)}{\sqrt{1+\frac{1}{n^4}}+\frac{1}{n^2}\sqrt{4+\frac{2}{n^2}}}\)
= lim \(-3n=-\infty\)
c) lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
= lim\(\frac{2+\frac{3}{n}}{\sqrt{9+\frac{3}{n^2}}-\sqrt[3]{\frac{2}{n}-8}}=\frac{2}{3+2}=\frac{2}{5}\)
Làm tính nhân: a. \(\left(3x^{2m-1}-\dfrac{3}{7}y^{3n-5}+x^{2m}y^{3m}-3y^2\right)8x^{3-2m}y^{6-3n}\)
b.\(\left(2x^{2n}+3x^{2n-1}\right)\left(x^{1-2n}-3x^{2-2n}\right)\)
a: \(=24x^{2m-1+3-2m}y^{6-3m}-\dfrac{24}{7}y^{3n-7+6-3n}\cdot x^{3-2m}+8x^{3-2m+2m}\cdot y^{6-3n+3m}-24x^{3-2m}y^{6-2n+2}\)
\(=24x^2y^{6-3m}-\dfrac{24}{7}x^{3-2m}\cdot y^{-1}+8x^3y^{-3n+3m+6}-24x^{3-2m}y^{-2n+8}\)
b: \(=2x^{2n+1-2n}-6x^{2n+2-2n}+3x^{2n-1+1-2n}-9x^{2n-1+2-2n}\)
\(=2x-6x^2+3-9x\)
\(=-6x^2-7x+3\)
Đề:Cho m,n là các số nguyên dương với \(n>1\).Đặt \(P=m^2n^2-4m+4n\)
Chứng minh rằng nếu P là số chính phương thì m=n
Giả sử \(m>n>1\)
Xét \(\left(mn^2-2\right)^2-n^2\left(m^2n^2-4m+4n\right)\)
\(=m^2n^4-4mn^2+4-mn^4+4mn^2-4n^3\)
\(=-4n^3+4< 0\) với \(\forall n>1\)
\(\Rightarrow\left(mn^2-2\right)^2< n^2\left(m^2n^2-4n+4n\right)\left(1\right)\)
Xét \(n^2\left(m^2n^2-4m+4n\right)-m^2n^4\)
\(=m^2n^4-4mn^2+4n^3-m^2n^4\)
\(=-4mn^2+4n^3\)
\(=-4n^2\left(m-n\right)< 0\) với \(\forall m>n>1\)
\(\Rightarrow n^2\left(m^2n^2-4m+4n\right)< m^2n^4\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\left(mn^2-2\right)^2< n^2\left(m^2n^2-4m+4n\right)< m^2n^4\)
\(\Rightarrow\left(\frac{mn^2-2}{n}\right)^2< P< \left(mn\right)^2\)
Xét \(\frac{mn^2-2}{n}-\left(mn-1\right)=\frac{n-2}{n}\ge0\) với \(\forall n\ge2\)
\(\Rightarrow\frac{mn^2-2}{n}\ge mn-1\)
\(\Rightarrow\left(mn-1\right)^2< P< \left(mn\right)^2\left(VL\right)\)
Kẹp giữa 2 số chính phương liên tiếp thì không tồn tại số chính phương nào.OK?
Giả sử \(m< n\)
\(\Rightarrow P>m^2n^2\left(3\right)\)
Xét \(m^2n^2-4m+4n-\left(mn+2\right)^2\)
\(=m^2n^2-4m+4n-m^2n^2-4mn-4\)
\(=n-m-mn-1=n\left(1-m\right)-m-1< 0\)
\(\Rightarrow P< \left(mn+2\right)^2\left(4\right)\)
Từ \(\left(3\right);\left(4\right)\Rightarrow\left(mn\right)^2< P< \left(mn+2\right)^2\)
Để P là số chính phương thì \(P=\left(mn+1\right)^2\)
\(\Rightarrow m^2n^2-4m+4n=m^2n^2+2mn+1\)
\(\Rightarrow-4m+4n-2mn=1\) quá VL
Với \(m=n\Rightarrow P=m^2n^2=\left(mn\right)^2\left(Lscp\right)\) cực kỳ HL:v
P/S:Ko chắc đâu nha.m thử làm bài 1 cấy.t cụng ra rồi nhưng coi cách m cho nó chắc:v Định dùng cách kẹp khác mà đề cho chặt quá:((
\(A\left(x\right)=Q\left(x\right)\left(x-1\right)+4\)(1)
\(A\left(x\right)=P\left(x\right)\left(x-3\right)+14\)(2)
\(A\left(x\right)=\left(x-1\right)\left(x-3\right)T\left(x\right)+F\left(x\right)\)(3)
Đặt : \(F\left(x\right)=ax+b\)
Với x=1 từ (1) và (3)
\(\hept{\begin{cases}A\left(1\right)=4\\A\left(1\right)=a+b\end{cases}}\)
\(\Rightarrow a+b=4\)(*)
Với x=3 từ (3) và (2)
\(\hept{\begin{cases}A\left(3\right)=14\\A\left(3\right)=3a+b\end{cases}}\)
\(\Rightarrow3a+b=14\)(**)
Từ (*) và (**)
\(\Rightarrow2a=10\Rightarrow a=5\Rightarrow b=-1\)
\(\Rightarrow F\left(x\right)=ax+b=5x-1\)
T lm r, ko bt có đúng ko:))
tính lim của lim\(\frac{4n^5-n+1}{\left(2n+1\right)\left(-n+1\right)\left(n^2+2\right)}\)
Tính lim \(\frac{\left(2n-1\right)\left(3n^2+2\right)^3}{-2n^5+4n^3-1}\)
\(=lim\frac{\left(2-\frac{1}{n}\right)\left(3n^{\frac{2}{3}}+\frac{2}{n^{\frac{4}{3}}}\right)^2}{-2+\frac{4}{n^2}-\frac{1}{n^5}}=\frac{\infty}{-2}=-\infty\)
CMR: với mọi số tự nhiên n :
a) \(\left(x+1\right)^{2n}-x^{2n}-2x-1\) chia hết cho \(x\left(x+1\right)\left(2x+1\right)\)
b) \(x^{4n+2}+2x^{2n+1}+1\) chia hết cho \(\left(x+1\right)^2\)
c) \(\left(x+1\right)^{4n+2}+\left(x-1\right)^{4n+2}\) chia hết cho \(x^2+1\)
1) cho hpt: \(\left\{{}\begin{matrix}x-3y=5-2m\\2x+y=3\left(m+1\right)\end{matrix}\right.\)
tìm m để hpt có nghiệm (\(x_0,y_0\)) t/m: \(x_0^2+y_0^2=9m\)
2) cho hpt: \(\left\{{}\begin{matrix}x+my=3m\\mx-y=m^2-2\end{matrix}\right.\)
tìm m để hpt có nghiệm duy nhất \(\left(x_0,y_0\right)\) t/m: \(x_0^2-2x_0-y_0>0\)
giúp mk vs mk cần gấp
Bài 1.
\(\left\{{}\begin{matrix}x-3y=5-2m\\2x+y=3\left(m+1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3y=5-2m\\6x+3y=9m+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}7x=7m+14\\x-3y=5-2m\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\m+2-3y=5-2m\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\-3y=-3m+3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=m+2\\y=m-1\end{matrix}\right.\)
\(x_0^2+y_0^2=9m\)
\(\Leftrightarrow\left(m+2\right)^2+\left(m-1\right)^2=9m\)
\(\Leftrightarrow m^2+4m+4+m^2-2m+1-9m=0\)
\(\Leftrightarrow2m^2-7m+5=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}m=1\\m=\dfrac{5}{2}\end{matrix}\right.\) ( Vi-ét )
Rút gọn biểu thức :
1. \(\frac{2^{4m}-2^{4n}}{2^{2n}+2^{2m}}\)
2. \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)
3. \(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}\)
4. \(\frac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+x\right)^2+\left(z-x\right)^2}\)
5. \(\frac{x^3+y^3+x^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
Please, help me!~~~ Pt2