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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
đặt \(A=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right)\)
\(\Rightarrow S=A.\left(\frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m}\right)=A.\frac{p}{m-n}+A.\frac{m}{n-p}+A.\frac{n}{p-m}\)
giờ ta xét từng hạng tử 1 nhé:
\(A.\frac{p}{m-n}=\left(\frac{m-n}{p}+\frac{n-p}{m}+\frac{p-m}{n}\right).\frac{p}{m-n}\)
\(=1+\frac{p}{m-n}.\left(\frac{n-p}{m}+\frac{p-m}{n}\right)\)
\(=1+\frac{p}{m-n}.\left(\frac{\left(n-p\right).n+m.\left(p-m\right)}{m.n}\right)\)
\(=1+\frac{p}{m-n}.\left(\frac{n^2-pn+m.p-m^2}{m.n}\right)\)
\(=1+\frac{p}{m-n}.\left(\frac{\left(n-m\right).\left(n+m\right)+p.\left(m-n\right)}{m.n}\right)\)
\(=1+\frac{p}{m-n}.\left(\frac{\left(p-m-n\right).\left(m-n\right)}{m.n}\right)\)
\(=1+\frac{p.\left(p-m-n\right)}{m.n}\)
\(=1+\frac{p^2-p.\left(m+n\right)}{m.n}\)
bây h ta sẽ sử dụng giả thiết \(m+n+p=0\Rightarrow m+n=-p\)
\(\Rightarrow A.\frac{p}{m-n}=1+\frac{p^2+p^2}{m.n}=1+\frac{2p^3}{m.n.p}\)
CM tương tự ta có: \(A.\frac{m}{n-p}=\frac{2m^3}{mnp}\) ; \(A.\frac{n}{p-m}=\frac{2n^3}{mnp}\)
\(\Rightarrow S=A.\left(\frac{p}{m-n}+\frac{m}{n-p}+\frac{n}{p-m}\right)=A.\frac{p}{m-n}+A.\frac{m}{n-p}+A.\frac{n}{p-m}=3+\frac{2\left(p^3+m^3+n^3\right)}{m.n.p}\)
\(m+n+p=0\Rightarrow\left(m+n+p\right).\left(m^2+p^2+n^2-mn-mp-np\right)=0\Leftrightarrow m^3+n^3+p^3-3mnp=0\)
\(\Leftrightarrow m^3+n^3+p^3=3mnp\)
\(S=3+\frac{2.3mnp}{mnp}=3+6=9\)
Vậy \(S=9\Leftrightarrow m+n+p=0\)
\(A=\dfrac{m^2+5m+n^2+5n+2mn-6}{m^2+6m+n^2+6n+2mn}\)
\(=\dfrac{\left(m+n\right)^2+5\left(m+n\right)-6}{\left(m+n\right)^2+6\left(m+n\right)}\)
\(=\dfrac{2013^2+5\cdot2013-6}{2013^2+6\cdot2013}=\dfrac{2012}{2013}\)
bài 1:
a) 4n+4+3n-6<19
<=> 7n-2<19
<=> 7n<21 <=> n< 3
b) n\(^2\) - 6n + 9 - n\(^2\) + 16\(\leq\)43
-6n+25\(\leq\)43
-6n\(\leq\)18
n\(\geq\)-3
\(\Leftrightarrow M=\left(\frac{x\left(x-2\right)}{2\left(x^2+4\right)}-\frac{2x^2}{4\left(2-x\right)+x^2\left(2-x\right)}\right)\left(\frac{x^2-x-2}{x^2}\right)\)
\(\Leftrightarrow M=\frac{x\left(x-2\right)\left(2-x\right)-4x^2}{2\left(x^2+4\right)\left(2-x\right)}.\frac{x^2-x-2}{x^2}\)
\(\Leftrightarrow M=\frac{-x\left(x^2-4x+4\right)-4x^2}{2\left(x^2+4\right)\left(2-x\right)}.\frac{x\left(x-2\right)+\left(x-2\right)}{x^2}\)
\(\Leftrightarrow M=\frac{x\left(2-x\right)\left(x+2\right)}{2\left(x^2+4\right)\left(2-x\right)}.\frac{\left(x+1\right)\left(x-2\right)}{x^2}\)hình như sai sai đề
Lời giải:
ĐKXĐ: $x\neq 2; x\neq 0$
a)
\(M=\left[\frac{x(x-2)}{2(x^2+4)}-\frac{2x^2}{(2-x)(x^2+4)}\right].\frac{x^2-x-2}{x^2}=\left[\frac{x(x-2)^2}{2(x^2+4)(x-2)}+\frac{4x^2}{2(x-2)(x^2+4)}\right].\frac{(x-2)(x+1)}{x^2}\)
\(=\frac{x(x-2)^2+4x^2}{2(x-2)(x^2+4)}.\frac{(x-2)(x+1)}{x^2}=\frac{x(x^2+4)}{2(x^2+4)(x-2)}.\frac{(x-2)(x+1)}{x^2}=\frac{x+1}{2x}\)
b)
Để $M$ nguyên thì $x+1\vdots 2x$
$\Rightarrow 2(x+1)\vdots 2x$
$\Rightarrow 2\vdots 2x\Rightarrow 1\vdots x$
Thay vào $M$ thấy $x=1$ thì $M=1$ là số nguyên dương.
c)
$M\geq -3\Leftrightarrow \frac{7x+1}{2x}\geq 0$
\(\left\{\begin{matrix} 7x+1\geq 0\\ 2x>0\end{matrix}\right.\) hoặc \(\left\{\begin{matrix} 7x+1\leq 0\\ 2x< 0\end{matrix}\right.\)
$\Rightarrow x>0$ hoặc $x\leq \frac{-1}{7}$
$\Rightarrow x=\pm 1$
Bài ấy dễ mà
\(\frac{m^3-n^3-3mn\left(m-2\right)}{m^2+n^2-2mn}\)
\(=\frac{m^3-n^3-3m^2n+3mn^2}{m^2-2mn+n^2}\)
\(=\frac{m^3-3m^2n-3mn^2-n^3}{m^2-2mn+n^2}=\frac{\left(m-n\right)^3}{\left(m-n\right)^2}=m-n\)
Thay m = 6,75 , n = -3,25 ta có :
6,75 - ( - 3,25 ) = 6,75 + 3,25 = 10
Vậy giá trị biểu thức trên bằng 10 khi m = 6,75 ; n = -3,25