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Mk muốn làm giúp bạn lắm chứ nhưng mà khổ lỗi mk mới học lớp 6 . Xin lỗi bn
bài 2 gợi ý từ hdt (x+y+z)^3=x^3+y^3+z^3+3(x+y)(y+z)(z+x)
VT (ở đề bài) = a+b+c
<=>....<=>3[căn bậc 3(a)+căn bậc 3(b)].[căn bậc 3(b)+căn bậc 3(c)].[căn bậc 3(c)+căn bậc 3 (a)]=0
từ đây rút a=-b,b=-c,c=-a đến đây tự giải quyết đc r
a, \(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}.\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)
\(=\left(\frac{1}{y-1}-\frac{y}{\left(1-y\right)\left(1+y+y^2\right)}.\frac{y^2+y+1}{y+1}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)
\(=\left(\frac{1}{y-1}+\frac{y\left(y^2+y+1\right)}{\left(y+1\right)^2\left(y^2+y+1\right)}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)
\(=\left(\frac{1}{y-1}+\frac{y}{\left(y+1\right)^2}\right):\frac{1}{\left(y-1\right)\left(x+1\right)}\)
\(=\left(\frac{\left(y+1\right)^2+y\left(y-1\right)}{\left(y-1\right)\left(y+1\right)^2}\right).\frac{\left(y-1\right)\left(y+1\right)}{1}=\frac{y^2+2y+1+y^2-y}{y+1}=\frac{2y^2+y+1}{y+1}\)
b, Thay y = 1/2 ta có :
\(\frac{2.\left(\frac{1}{2}\right)^2+\frac{1}{2}+1}{\frac{1}{2}+1}=\frac{\frac{1}{2}+\frac{1}{2}+\frac{2}{2}}{\frac{1}{2}+\frac{2}{2}}=\frac{\frac{5}{2}}{\frac{3}{2}}=\frac{5}{12}\)
cau a),b),c) ban dat mau chung roi khu mau ma lam la duoc ma
Tuy học lớp 6 ................. cơ mừ thấy mí bài nỳ dễ quá >.<
a) ĐKXĐ : \(y\ne\pm1\)
\(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}.\frac{y^2+y+1}{y+1}\right)\div\frac{1}{y^2-1}\)
\(=\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y^2+y+1\right)}.\frac{y^2+y+1}{y+1}\right)\div\frac{1}{y^2-1}\)
\(=\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y+1\right)}\right)\div\frac{1}{y^2-1}\)
\(=\frac{y+1+y}{\left(y-1\right)\left(y+1\right)}\div\frac{1}{\left(y-1\right)\left(y+1\right)}\)
\(=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}.\left(y-1\right)\left(y+1\right)\)
\(=2y+1\)
Vậy \(N=2y+1\)khi \(y\ne\pm1\)
b) Với \(y=\frac{1}{2}\); phương trình N trở thành :
\(N=2.\frac{1}{2}+1=2\)
Vậy N=2 khi \(y=\frac{1}{2}\)
c) Để N luôn dương
\(\Leftrightarrow2y+1>0\)
\(\Leftrightarrow2y>-1\)
\(\Leftrightarrow y>\frac{-1}{2}\)
Kết hợp ĐKXĐ ta có : \(y>\frac{-1}{2};y\ne\pm1\)
Vậy N luôn dương khi \(y>\frac{-1}{2};y\ne\pm1\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)
\(N=\frac{\left(x+2\right)^2}{x}.\left(1-\frac{x^2}{x+2}\right)-\frac{x^2+6x+4}{x}\)
\(N=\frac{\left(x+2\right)^2}{x}.\frac{x+2-x^2}{x+2}-\frac{x^2+6x+4}{x}\)
\(N=\frac{\left(x+2\right)\left(x+2-x^2\right)-x^2-6x-4}{x}\)
\(N=\frac{x^2+2x-x^3+2x+4-2x^2-x^2-6x-4}{x}\)
\(N=\frac{-x^3-2x^2-2x}{x}\)
\(N=\frac{-x\left(x^2+2x+2\right)}{x}\)
\(N=-\left(x^2+2x+2\right)\)
b) \(N=-\left(x^2+2x+2\right)\)
\(\Leftrightarrow N=-\left(x^2+2x+1+1\right)\)
\(\Leftrightarrow N=-\left(x+1\right)^2-1\le-1\)
Max N = -1 \(\Leftrightarrow x=-1\)
Vậy .......................
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
\(\frac{1}{n\left(n+1\right)}=\frac{a}{n}+\frac{b}{n+1}\)
\(\Leftrightarrow\frac{\left(n+1\right)-n}{n\left(n+1\right)}=\frac{a\left(n+1\right)+bn}{n\left(n+1\right)}\)
Đến đây đồng nhất ta được : \(\hept{\begin{cases}a=1\\b=-1\end{cases}}\)