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Ta có : \(n^4+4=\left[\left(n-1\right)^2+1\right]\left[\left(n+1\right)^2+1\right]\)
Do đó :
\(M=\frac{1\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\frac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.\frac{\left(8^2+1\right)\left(10^2+1\right)}{\left(10^2+1\right)\left(12^2+1\right)}...\frac{\left(16^2+1\right)\left(18^2+1\right)}{\left(18^2+1\right)\left(20^2+1\right)}\)
\(M=\frac{1}{20^2+1}=\frac{1}{401}\)
a,\(=\left(\frac{3}{5}x+\frac{2}{7}y\right)^2=\left(\frac{3}{5}.5+\frac{2}{7}.\left(-7\right)\right)^2=0\)
\(b,=\left(\frac{5}{4}u^2v+\frac{2}{25}v^2\right)^2=\left(\frac{5}{4}.\left(\frac{2}{5}\right)^2.5+\frac{2}{25}.5^2\right)^2=3^2=9\)
bài 1) Đặt \(B=\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\)
Ta có: \(A=B.\left(\frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m}\right)=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}\)
\(B.\frac{p}{m-n}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{p}{m-n}=\frac{m-n}{p}.\frac{p}{m-n}+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}\)
\(=1+\frac{n-p}{m}.\frac{p}{m-n}+\frac{p-m}{n}.\frac{p}{m-n}=1+\frac{p}{m-n}.\left(\frac{n-p}{m}+\frac{p-m}{n}\right)\)
\(=1+\frac{p}{m-n}.\left[\frac{\left(n-p\right).n}{mn}+\frac{\left(p-m\right).m}{mn}\right]=1+\frac{p}{m-n}.\frac{n^2-np+pm-m^2}{mn}\)
\(=1+\frac{p}{m-n}.\frac{\left(m-n\right).\left(p-m-n\right)}{mn}=1+\frac{p.\left(m-n\right).\left(p-m-n\right)}{\left(m-n\right).mn}=1+\frac{p.\left(p-m-n\right)}{mn}\)
\(=1+\frac{p^2-pm-pn}{mn}=1+\frac{p^2-p.\left(m+n\right)}{mn}\)
Vì m+n+p=0=>m+n=-p
\(=>B.\frac{p}{m-n}=1+\frac{p^2-p.\left(-p\right)}{mn}=1+\frac{2p^2}{mn}=1+\frac{2p^3}{mnp}\left(1\right)\)
\(B.\frac{m}{n-p}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{m}{n-p}=\frac{m-n}{p}.\frac{m}{n-p}+\frac{n-p}{m}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}\)
\(=1+\frac{m-n}{p}.\frac{m}{n-p}+\frac{p-m}{n}.\frac{m}{n-p}=1+\frac{m}{n-p}.\left(\frac{m-n}{p}+\frac{p-m}{n}\right)\)
\(=1+\frac{m}{n-p}.\left[\frac{\left(m-n\right).n}{np}+\frac{\left(p-m\right).p}{np}\right]=1+\frac{m}{n-p}.\frac{mn-n^2+p^2-mp}{np}\)
\(=1+\frac{m}{n-p}.\frac{\left(n-p\right).\left(m-n-p\right)}{np}=1+\frac{m.\left(n-p\right).\left(m-n-p\right)}{\left(n-p\right).np}=1+\frac{m.\left(m-n-p\right)}{np}\)
\(=1+\frac{m^2-mn-mp}{np}=1+\frac{m^2-m\left(n+p\right)}{np}=1+\frac{m^2-m.\left(-m\right)}{np}=1+\frac{2m^2}{np}=1+\frac{2m^3}{mnp}\left(2\right)\) (vì m+n+p=0=>n+p=-m)
\(B.\frac{n}{p-m}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{n}{p-m}=\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}+\frac{p-m}{n}.\frac{n}{p-m}\)
\(=1+\frac{m-n}{p}.\frac{n}{p-m}+\frac{n-p}{m}.\frac{n}{p-m}=1+\frac{n}{p-m}.\left(\frac{m-n}{p}+\frac{n-p}{m}\right)\)
\(=1+\frac{n}{p-m}.\left[\frac{\left(m-n\right).m}{pm}+\frac{\left(n-p\right).p}{pm}\right]=1+\frac{n}{p-m}.\frac{m^2-mn+np-p^2}{pm}\)
\(=1+\frac{n}{p-m}.\frac{\left(p-m\right).\left(n-p-m\right)}{pm}=1+\frac{n.\left(p-m\right).\left(n-p-m\right)}{\left(p-m\right).pm}=1+\frac{n.\left(n-p-m\right)}{pm}\)
\(=1+\frac{n^2-np-mn}{pm}=1+\frac{n^2-n\left(p+m\right)}{pm}=1+\frac{n^2-n.\left(-n\right)}{pm}=1+\frac{2n^2}{pm}=1+\frac{2n^3}{mnp}\left(3\right)\) (vì m+n+p=0=>p+m=-n)
Từ (1),(2),(3) suy ra :
\(A=B.\frac{p}{m-n}+B.\frac{m}{n-p}+B.\frac{n}{p-m}=\left(1+\frac{2p^3}{mnp}\right)+\left(1+\frac{2m^3}{mnp}\right)+\left(1+\frac{2n^3}{mnp}\right)\)
\(=3+\frac{2p^3}{mnp}+\frac{2m^3}{mnp}+\frac{2n^3}{mnp}=3+\frac{2.\left(m^3+n^3+p^3\right)}{mnp}\)
*Tới đây để tính được m3+n3+p3,ta cần CM được bài toán phụ sau:
Đề: Cho m+n+p=0.CMR: \(m^3+n^3+p^3=3mnp\)
Từ m+n+p=0=>m+n=-p
Ta có: \(m^3+n^3+p^3=\left(m+n\right)^3-3m^2n-3mn^2+p^3=-p^3-3mn\left(m+n\right)+p^3\)
\(=-3mn\left(m+n\right)=-3mn.\left(-p\right)=3mnp\)
Vậy ta đã CM được bài toán phụ
*Trở lại bài toán chính: \(A=3+\frac{2.3mnp}{mnp}=3+\frac{6mnp}{mnp}=3+6=9\)
Vậy A=9
bài 2)
a)Nhận thấy các thừa số của A đều có dạng tổng quát sau:
\(n^3+1=n^3+1^3=\left(n+1\right)\left(n^2-n+1\right)=\left(n+1\right).\left(n^2-n+\frac{1}{4}+\frac{3}{4}\right)\)
\(=\left(n+1\right).\left(n^2-2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n+1\right).\left[\left(n-\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]\)
\(n^3-1=n^3-1^3=\left(n-1\right)\left(n^2+n+1\right)=\left(n-1\right).\left(n^2+n+\frac{1}{4}+\frac{3}{4}\right)\)
\(=\left(n-1\right).\left(n^2+2.n.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\right)=\left(n-1\right).\left[\left(n+\frac{1}{2}\right)^2+\frac{3}{4}\right]=\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]\)
suy ra \(\frac{n^3+1}{n^3-1}=\frac{\left(n+1\right).\left[\left(n-0,5\right)^2+0,75\right]}{\left(n-1\right).\left[\left(n+0,5\right)^2+0,75\right]}\)
Do đó: \(\frac{2^3+1}{2^3-1}=\frac{\left(2+1\right).\left[\left(2-0,5\right)^2+0,75\right]}{\left(2-1\right).\left[\left(2+0,5\right)^2+0,75\right]}=\frac{3.\left(1,5^2+0,75\right)}{1.\left(2,5^2+0,75\right)}\)
\(\frac{3^3+1}{3^3-1}=\frac{\left(3+1\right).\left[\left(3-0,5\right)^2+0,75\right]}{\left(3-1\right).\left[\left(3+0,5\right)^2+0,75\right]}=\frac{4.\left(2,5^2+0,75\right)}{2.\left(3,5^2+0,75\right)}\)
...........................
\(\frac{10^3+1}{10^3-1}=\frac{\left(10+1\right).\left[\left(10-0,5\right)^2+0,75\right]}{\left(10-1\right).\left[\left(10+0,5\right)^2+0,75\right]}=\frac{11.\left(9,5^2+0,75\right)}{9.\left(10,5^2+0,75\right)}\)
\(=>A=\frac{3\left(1,5^2+0,75\right).4\left(2,5^2+0,75\right)........11.\left(9,5^2+0,75\right)}{1\left(2,5^2+0,75\right).2.\left(3,5^2+0,75\right)........9\left(10,5^2+0,75\right)}=\frac{3.4........11}{1.2......9}.\frac{1,5^2+0,75}{10,5^2+0,75}\)
\(=\frac{10.11}{2}.\frac{1}{37}=\frac{2036}{37}\)
Vậy A=2036/37
b) có thể ở chỗ 1+1/4 bn nhầm,phải là \(1^4+\frac{1}{4}\) ,mà chắc cũng chẳng sao,vì 14=1 mà
Nhận thấy các thừa số của B có dạng tổng quát:
\(n^4+\frac{1}{4}=n^4+n^2+\frac{1}{4}-n^2=\left(n^2\right)^2+2.n^2.\frac{1}{2}+\frac{1}{4}-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)
\(=\left(n^2+\frac{1}{2}-n\right)\left(n^2+\frac{1}{2}+n\right)\)
\(B=\frac{\left(1^2+\frac{1}{2}-1\right).\left(1^2+\frac{1}{2}+1\right).\left(3^2+\frac{1}{2}+3\right).\left(3^2+\frac{1}{2}-3\right)..........\left(9^2+\frac{1}{2}-9\right).\left(9^2+\frac{1}{2}+9\right)}{\left(2^2+\frac{1}{2}-2\right).\left(2^2+\frac{1}{2}+2\right).\left(4^2+\frac{1}{2}-4\right).\left(4^2+\frac{1}{2}+4\right)......\left(10^2+\frac{1}{2}-10\right).\left(10^2+\frac{1}{2}+10\right)}\)
Mặt khác,ta cũng có: \(\left(a+1\right)^2-\left(a+1\right)+\frac{1}{2}=a^2+2a+1-a-1+\frac{1}{2}=a^2+a+\frac{1}{2}\)
Suy ra \(B=\frac{1^2+\frac{1}{2}-1}{10^2+\frac{1}{2}+10}=\frac{1}{221}\)
Vậy B=1/221
a) Ta có: \(\frac{3x-2}{6}-\frac{4-3x}{18}=\frac{4-x}{9}\)
\(\Leftrightarrow\frac{3\left(3x-2\right)}{18}-\frac{4-3x}{18}-\frac{2\left(4-x\right)}{18}=0\)
\(\Leftrightarrow9x-6-4+3x-\left(8-2x\right)=0\)
\(\Leftrightarrow12x-10-8+2x=0\)
\(\Leftrightarrow10x-18=0\)
\(\Leftrightarrow10x=18\)
hay \(x=\frac{9}{5}\)
Vậy: \(x=\frac{9}{5}\)
b) Ta có: \(\frac{2+3x}{6}-x+2=\frac{x-7}{9}\)
\(\Leftrightarrow\frac{3\left(2+3x\right)}{18}-\frac{18x}{18}+\frac{36}{18}-\frac{2\left(x-7\right)}{18}=0\)
\(\Leftrightarrow6+9x-18x+36-\left(2x-14\right)=0\)
\(\Leftrightarrow42-9x-2x+14=0\)
\(\Leftrightarrow56-11x=0\)
\(\Leftrightarrow11x=56\)
hay \(x=\frac{56}{11}\)
Vậy: \(x=\frac{56}{11}\)
c) ĐKXĐ: x∉{3;-3}
Ta có: \(\frac{6-x}{x^2-9}+\frac{2}{x+3}=\frac{-5}{x-3}\)
\(\Leftrightarrow\frac{6-x}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{-5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow6-x+2x-6=-5x-15\)
\(\Leftrightarrow x+5x+15=0\)
\(\Leftrightarrow6x=-15\)
hay \(x=\frac{-5}{2}\)(tm)
Vậy: \(x=\frac{-5}{2}\)
d) Ta có: \(\left(5x+2\right)\left(x^2-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x^2-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-2\\x^2=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{5}\\x=\pm\sqrt{7}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-2}{5};\sqrt{7};-\sqrt{7}\right\}\)
e) ĐKXĐ: x∉{4;-4}
Ta có: \(\frac{3}{x-4}+\frac{5x-2}{x^2-16}=\frac{4}{x+4}\)
\(\Leftrightarrow\frac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{5x-2}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Leftrightarrow3x+12+5x-2-\left(4x-16\right)=0\)
\(\Leftrightarrow8x+10-4x+16=0\)
\(\Leftrightarrow4x+26=0\)
\(\Leftrightarrow4x=-26\)
hay \(x=\frac{-13}{2}\)(tm)
Vậy: \(x=\frac{-13}{2}\)
Bạn viết biểu thức A ra đi rồi bọn mình mới làm được chứ -.-
Đk : \(x\ne\pm3\)
Để B>A
\(\Leftrightarrow\frac{3}{x+3}>4\)
Rõ ràng: \(x+3>0\)
\(\Rightarrow\frac{3}{x+3}>4\)
\(\Leftrightarrow3>4\left(x+3\right)\)
\(\Leftrightarrow3>4x+12\)
\(\Leftrightarrow-9>4x\)
\(\Leftrightarrow x< \frac{-9}{4}\)
KL: \(x\in Z,x< \frac{-9}{4},x\ne\pm3\)
Công thức tổng quát ''mở'' cho bài toán trên được hình thành trên cơ sở phân tích thành nhân tử và được phát biểu như sau:
\(a^4+4=\left(a^2-2a+2\right)\left(a^2+2a+2\right)\)
Khi đó, biểu thức \(A\) trở thành:
\(A=\frac{\left(1^2-2+2\right)\left(1^2+2+2\right)\left(5^2-2.5+2\right)\left(5^2+2.5+2\right)...\left(17^2-2.17+2\right)\left(17^2+2.17+2\right)}{\left(3^2-2.3+2\right)\left(3^2+2.3+2\right)\left(7^2-2.7+2\right)\left(7^2+2.7+2\right)...\left(19^2-2.19+2\right)\left(19^2+2.19+2\right)}\)
\(A=\frac{\left(1^2-2+2\right)}{\left(19^2+2.19+2\right)}=\frac{1}{401}\)