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\(\dfrac{x+1}{65}+\dfrac{x+3}{63}\) = \(\dfrac{x+5}{61}\) + \(\dfrac{x+7}{59}\)
<=> \(\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1\) = \(\dfrac{x+5}{61}\) + 1 + \(\dfrac{x+7}{59}\) + 1
<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) = \(\dfrac{x+66}{61}\) + \(\dfrac{x+66}{59}\)
<=> \(\dfrac{x+66}{65}+\dfrac{x+66}{63}\) - \(\dfrac{x+66}{61}\) - \(\dfrac{x+66}{59}\) = 0
<=> (x + 66) . (\(\dfrac{1}{65}+\dfrac{1}{63}+\dfrac{1}{61}+\dfrac{1}{59}\)) = 0
<=> x + 66 = 0
<=> x = -66
a.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+2}=a\\\sqrt[3]{x-2}=b\end{matrix}\right.\) ta được:
\(2a^2-b^2=ab\)
\(\Leftrightarrow\left(a-b\right)\left(2a+b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a=-b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\8a^3=-b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=x-2\left(vô-nghiệm\right)\\8\left(x+2\right)=-\left(x-2\right)\end{matrix}\right.\)
\(\Leftrightarrow x=-\dfrac{14}{9}\)
b.
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\)
\(\Rightarrow a^2+4b^2=5ab\)
\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=4b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}a^3=b^3\\a^3=64b^3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}65+x=65-x\\65+x=64\left(65-x\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
a) Đặt \(\sqrt[3]{65+x}=a;\sqrt[3]{65-x}=b\)
Nhận xét x = 65 không phải là nghiệm. Xét x khác 65 thì \(b\ne0\)
PT \(\Leftrightarrow a^2+b^2-5ab=0\)
\(\Leftrightarrow\left(\frac{a}{b}\right)^2-5\left(\frac{a}{b}\right)+1=0\Leftrightarrow t^2-5t+1=0\left(\text{đặt }t=\frac{a}{b}\right)\)
Hình như chị ghi đề sai, số quá xấu:((
a/ Nghiệm xấu quá
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{65+x}=a\\\sqrt[3]{65-x}=b\end{matrix}\right.\) ta được:
\(a^2+b^2=5ab\Leftrightarrow a^2-5ab+b^2=0\)
\(\Leftrightarrow\left(a-\frac{5+\sqrt{21}}{2}b\right)\left(a-\frac{5-\sqrt{21}}{2}b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{5+\sqrt{21}}{2}b\\a=\frac{5-\sqrt{21}}{2}b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{65+x}=\frac{5+\sqrt{21}}{2}\sqrt[3]{65-x}\\\sqrt[3]{65+x}=\frac{5-\sqrt{21}}{2}\sqrt[3]{65-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}65+x=\left(\frac{5+\sqrt{21}}{2}\right)^3\left(65-x\right)\\65+x=\left(\frac{5-\sqrt{21}}{2}\right)^3\left(65-x\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(56+12\sqrt{21}\right)x=65\left(54+12\sqrt{21}\right)\\\left(56-12\sqrt{21}\right)x=65\left(54-12\sqrt{21}\right)\end{matrix}\right.\) \(\Rightarrow x=...\)
b/ \(\Leftrightarrow\sqrt[3]{x-5}+\sqrt[3]{2x-1}=\sqrt[3]{3x+2}-2\)
\(\Leftrightarrow3x-6+3\sqrt[3]{\left(x-5\right)\left(2x-1\right)}\left(\sqrt[3]{3x+2}-2\right)=3x-6-6\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)\)
\(\Leftrightarrow\sqrt[3]{\left(x-5\right)\left(2x-1\right)}\left(\sqrt[3]{3x+2}-2\right)=-2\sqrt[3]{3x+2}\left(\sqrt[3]{3x+2}-2\right)\)
\(\Leftrightarrow\left(\sqrt[3]{3x+2}-2\right)\left(\sqrt[3]{\left(x-5\right)\left(2x-1\right)}+2\sqrt[3]{3x+2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=8\Rightarrow x=2\\\left(x-5\right)\left(2x-1\right)=-8\left(3x+2\right)\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow2x^2-13x+21=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\frac{7}{2}\end{matrix}\right.\)
Bài 1:
\(A=\frac{1}{\left(1+2\right)}+\frac{1}{\left(1+2+3\right)}+\frac{1}{\left(1+2+3+4\right)}\)\(+\frac{1}{\left(1+2+3+4+5\right)}+...+\)\(\frac{1}{\left(1+2+3+...+10\right)}\)
\(A=\frac{1}{3}+\frac{1}{6}+....+\frac{1}{55}\)
\(A=2\left(\frac{1}{6}+\frac{1}{12}+....+\frac{1}{110}\right)\)
\(A=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}\right)\)
\(A=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\right)\)
\(A=2.\left(\frac{1}{2}-\frac{1}{11}\right)\)
\(A=\frac{9}{11}\)
Bài 2 :
2) Tử số = 11 x 13 x 15 + 3 x 3 x 3 x 11 x 13 x 15 + 5 x 5 x 5 x 11 x 13 x 15 + 9 x 9 x 9 x 11 x 13 x 15
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) x 11 x 13 x 15 = (1+27+125+ 729) x 11 x 13 x 15
Mẫu số = 11 x 13 x 17 + 3 x 3 x 3 x 13 x 15 x 19 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17 lớn hơn 11 x 13 x 15 + 3 x 3 x 3 x 13 x 15 x 17 + 5 x 5 x 5 x 13 x 15 x 17 + 9 x 9 x 9 x 13 x 15 x 17
= (1 + 3 x 3 x 3 + 5 x 5 x 5 + 9 x 9 x9) 13 x 15 x 17 = (1+27+125+729) x 13 x 15 x 17
\(\Rightarrow A< \frac{\left(1+27+125+729\right)\times11\times13\times15}{\left(1+27+125+729\right)\times13\times15\times17}\)
\(=\frac{11}{17}\)
\(=\frac{1111}{1717}=B\)
Vậy \(A=B\)
\(65.x-\frac{1}{3}.x=\frac{2}{5}\)
\(\left(65-\frac{1}{3}\right).x=\frac{2}{5}\)
\(\frac{194}{3}.x=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{194}{3}\)
\(x=\frac{3}{485}\)
. là nhân nha
65 x X - \(\frac{1}{3}\)x X = \(\frac{2}{5}\)
( 65 - \(\frac{1}{3}\)) x X = \(\frac{2}{5}\)
\(\frac{194}{3}\)x X = \(\frac{2}{5}\)
X = \(\frac{2}{5}:\frac{194}{3}\)
X = \(\frac{291}{5}\)
Nếu mình đúng thì các bạn k mình nhé
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}+\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)
\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)\)
\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}+\dfrac{x+66}{61}+\dfrac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]\)\(=0\)
Do \(\dfrac{1}{65}< \dfrac{1}{63}< \dfrac{1}{61}< \dfrac{1}{59}\)
\(\Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)< 0\)
Vậy để \(\left(x+66\right).\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)
\(\Leftrightarrow x+66=0\)
\(\Leftrightarrow x=-66\)
Vậy \(x\in\left\{-66\right\}\)
( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ( x + 4 ) + ( x + 5 ) = 65
( x + x + x + x + x ) + ( 1 + 2 + 3 + 4 = 5 ) = 65
x * 5 +15 = 65
x * 5 = 65 - 15
x * 5 = 50
x = 50 : 5
x = 10
( X + 1 ) + ( X + 2 ) + ( X + 3 ) + ( X + 4 ) + ( X + 5 ) = 65
X x 5 + ( 1 + 2 + 3 + 4 + 5 ) = 65
X x 5 + 15 = 65
X x 5 = 65 - 15
X x 5 = 50
X = 50 : 5
X = 10
\(\left(x+3\right)^3-1=-65\)
\(\Leftrightarrow\left(x+3\right)^3=-64\)
\(\Leftrightarrow x+3=-8\)
\(\Leftrightarrow x=-11\)