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\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)
\(\Leftrightarrow\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
Mà \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\ne0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy ..
\(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}=\dfrac{x+1}{5}+\dfrac{x+1}{6}\)
=> \(\dfrac{x+1}{2}+\dfrac{x+1}{3}+\dfrac{x+1}{4}-\dfrac{x+1}{5}-\dfrac{x+1}{6}\)= 0
(x + 1).(\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\)) = 0
Ta thấy \(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}\) > 0
=> x + 1 = 0
x = 0 - 1
x = -1
Giải:
a) \(-1313x^2y.2xy^3\)
\(=\left(-1313.2\right)\left(x^2.x\right)\left(y.y^3\right)\)
\(=-2626x^3y^4\)
Bậc của đơn thức là: \(3+4=7\)
b) \(1414x^3y.\left(-2x^3y^5\right)\)
\(=\left[1414.\left(-2\right)\right]\left(x^3.x^3\right)\left(y.y^5\right)\)
\(=-2828x^6y^6\)
Bậc của đơn thức là: \(6+6=12\).
Chúc bạn học tốt!!!
a) -x2y. 2xy3 = -2x3y4. Đơn thức có bậc là 7
b) x3y. (-2x3y5) = -2x6y6. Đơn thức có bậc là 12
>> Mình không chép lại đề bài nhé ! <<
Cách 1 :
\(A=\left(\dfrac{36-4+3}{6}\right)-\left(\dfrac{30+10-9}{6}\right)-\left(\dfrac{18-14+15}{6}\right)=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}=-\dfrac{15}{6}=-\dfrac{5}{2}\)
Cách 2 :
\(A=6-\dfrac{2}{3}+\dfrac{1}{2}-5+\dfrac{5}{3}-\dfrac{3}{2}-3-\dfrac{7}{3}+\dfrac{5}{2}\)
\(A=\left(6-5-3\right)-\left(\dfrac{2}{3}+\dfrac{5}{3}-\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\right)\)
\(A=-2-0-\dfrac{1}{2}=-\dfrac{5}{2}\)
Cách 1 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=\left(\dfrac{36}{6}-\dfrac{4}{6}+\dfrac{3}{6}\right)-\left(\dfrac{30}{6}+\dfrac{10}{6}-\dfrac{9}{6}\right)-\left(\dfrac{18}{6}-\dfrac{14}{6}+\dfrac{15}{6}\right)\)
\(=\dfrac{35}{6}-\dfrac{31}{6}-\dfrac{19}{6}\)
\(=-\dfrac{5}{2}\)
Cách 2 :
\(\left(6-\dfrac{2}{3}+\dfrac{1}{2}\right)-\left(5+\dfrac{5}{3}-\dfrac{3}{2}\right)-\left(3-\dfrac{7}{3}+\dfrac{5}{2}\right)\)
\(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=\left(6-5-3\right)+\left(\dfrac{-2}{3}+\dfrac{-5}{3}+\dfrac{7}{3}\right)+\left(\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{-5}{2}\right)\)
\(=\left(-2\right)+0+\dfrac{-1}{2}\)
\(=\dfrac{-5}{2}\)
a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (1)
\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\dfrac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)(2)
Từ (1) và (2) suy ra: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
b.M = \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{50^2}\right)\)
= \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{2499}{2500}\)
= \(\dfrac{1.3.2.4.3.5...49.51}{2^2.3^2.4^2...50^2}\)
\(\dfrac{51}{2.50}=\dfrac{51}{100}\)
Lời giải:
a)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow \left(\frac{a}{b}\right)^2=\left(\frac{b}{d}\right)^2=\frac{(a+c)^2}{(b+d)^2}(1)\)
Mặt khác, \(\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}(2)\) (áp dụng tính chất dãy tỉ số bằng nhau)
Từ \((1),(2)\Rightarrow \frac{(a+c)^2}{(b+d)^2}=\frac{a^2+c^2}{b^2+d^2}\)
b) Vì \(1-\frac{1}{2^2};1-\frac{1}{3^2};...;1-\frac{1}{50^2}<1\) nên:
\(\left\{\begin{matrix} \left \{ 1-\frac{1}{2^2} \right \}=1-\frac{1}{2^2}\\ \left \{ 1-\frac{1}{3^2} \right \}=1-\frac{1}{3^2}\\ ....\\ \left \{ 1-\frac{1}{50^2} \right \}=1-\frac{1}{50^2}\end{matrix}\right.\)
\(\Rightarrow M=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{50^2}\right)\)
\(\Leftrightarrow M=\frac{(2^2-1)(3^2-1)(4^2-1)....(50^2-1)}{(2.3....50)^2}\)
\(\Leftrightarrow M=\frac{[(2-1)(3-1)...(50-1)][(2+1)(3+1)...(50+1)]}{(2.3.4...50)^2}\)
\(\Leftrightarrow M=\frac{(2.3...49)(3.4.5...51)}{(2.3.4...50)^2}=\frac{(2.3.4...49)^2.50.51}{2.(2.3....49)^2.50^2}=\frac{50.51}{2.50^2}=\frac{51}{100}\)
\(VT=\dfrac{a+c}{a+b}+\dfrac{b+d}{b+c}+\dfrac{c+a}{c+d}+\dfrac{d+b}{d+a}\)
\(=\left(a+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{c+d}\right)+\left(b+d\right)\left(\dfrac{1}{b+c}+\dfrac{1}{d+a}\right)\)
Ap dụng \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y} \left(\forall x,y>0\right)\)
Ta có: \(VT\ge\left(a+c\right).\dfrac{4}{a+b+c+d}+\left(b+d\right).\dfrac{4}{a+b+c+d}\)
\(=\dfrac{4\left(a+b+c+d\right)}{\left(a+b+c+d\right)}=4\left(ĐPCM\right)\)
Ta có:\(\left(-5a^2b^4c^6\right)^7-\left(9a^3bc^5\right)^8=0\)
\(\left(-5\right)^7a^{14}b^{28}c^{42}-9^8a^{24}b^8c^{40}=0\)
Vì \(a^{14}b^{28}c^{42}\ge0\Rightarrow\left(-5\right)^7a^{14}b^{28}c^{42}\le0\)
\(a^{24}b^8c^{40}\ge0\Rightarrow9^8a^{24}b^8c^{40}\ge0\)
\(\Rightarrow\left(-5\right)^7a^{14}b^{28}c^{42}-9^8a^{24}b^8c^{40}\le0\)
Mà VP=0
Dấu "=" xảy ra khi
\(\left(-5\right)^7a^{14}b^{28}c^{42}=0\) và \(9^8a^{24}b^8c^{40}=0\)
\(\Rightarrow a=b=c=0\)
\(\Rightarrow A=a+b+c=0+0+0=0\)
Từ a/b=c/d⇒a/c=b/d
Áp dụng tính chất dãy tỉ số bằng nhau
a/c=b/d=a+b/c+d
⇒a^3/c^3=b^3/d^3=(a+b)^3/(c+d)^3 (1)
Từ a^3/c^3=b^3/d^3=a^3-b^3/c^3-d^3 (2)
Từ (1) và (2)
⇒(a+b)^3/(c+d)^3=a^3-b^3/c^3-d^3
b,
\(B=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)
\(\Rightarrow B=\frac{1}{1999.2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)
\(\Rightarrow B=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)
\(\Rightarrow B=\frac{-1997}{1999}-\frac{1}{2000}\)
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\left(1\right)\\ \Leftrightarrow1+\dfrac{a}{b}+\dfrac{b}{a}+1-4\ge0\\ \Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}-2\ge0\left(2\right)\)
Áp dụng t/c \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\) nên (2) luôn đúng.Do đó:(1) đúng
Vậy...(đpcm)
\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\) ( đúng)
\(\Rightarrow1+\dfrac{a}{b}+\dfrac{b}{a}+1-4\ge0\)
\(\Rightarrow\dfrac{a}{b}+\dfrac{b}{a}-2\ge0\) ( đúng)
Ta áp dụng tính chất \(\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
Vậy .....
Chúc bạn học tốt!