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\(\frac{5x}{1.6}+\frac{5x}{6.11}+\frac{5x}{11.16}+\frac{5x}{16.21}=\frac{1}{25}\)
\(x\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+\frac{5}{16.21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}\right)=\frac{1}{25}\)
\(x\left(1-\frac{1}{21}\right)=\frac{1}{25}\)
\(\frac{20}{21}x=\frac{1}{25}\)
\(x=\frac{1}{25}:\frac{20}{21}=.....\)
\(7.3^x+20.3^x=3^{25}\)
\(3^x.\left(7+20\right)=3^{25}\)
\(3^x.3^3=3^{25}\)
\(3^x=3^{22}\)
\(\Rightarrow x=22\)
7.3x+20.3x=325
\(\Leftrightarrow\)3x . ( 7 + 20 ) = 325
\(\Leftrightarrow\)3x . 27 = 325
\(\Leftrightarrow\)3x . 33 = 325
\(\Leftrightarrow\)3x = 325 : 33
\(\Leftrightarrow\)3x = 325 - 3
\(\Leftrightarrow\)3x = 323
\(\Leftrightarrow\)x = 23
Vậy x = 23
a) p= 2011.2019= 2011.(2015+4)= 2011.2015+2011.4
q= 2015.2015=(2011+4).2015= 2015.2011+2015.4
Do đó p<q
b) (x-25):15=20
x-25 = 20x15
x-25 =300
x =300+25
x =325
c) Đặt A = 2 + 4 + 6 + 8 + .....+ 2500
Số số hạng của A là :
(2500 - 2) : 2 + 1 = 1250 (số)
Tổng A là :
(2500 + 2) x 1250 : 2 = 1563750
Thay A vào ta có : x(x + 1) = 1563750
=> x(x + 1) = 1250 x 1251
=> x = 1250
Vậy x = 1250
Câu 2:
b: \(\left(x-25\right):15=20\)
=>\(x-25=20\cdot15=300\)
=>x=300+25=325
Câu 1:
\(p=2011\cdot2019\)
\(=\left(2015-4\right)\left(2015+4\right)\)
\(=2015\cdot2015-4\cdot4\)
=q-16
=>p<q
Câu 3:
\(x\left(x+1\right)=2+4+6+8+10+\cdots+2500\)
=>\(x\left(x+1\right)=2\left(1+2+\cdots+1250\right)\)
=>\(x\left(x+1\right)=2\cdot\frac{1250\cdot1251}{2}\)
=>\(x\left(x+1\right)=1250\cdot1251\)
=>\(x^2+x-1250\cdot1251=0\)
=>(x+1251)(x-1250)=0
=>\(\left[\begin{array}{l}x+1251=0\\ x-1250=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-1251\\ x=1250\end{array}\right.\)
Ta có:
x(x-5.12)=0
TH1:x=0
TH2:x-5.12 hay x-60=0
=>x=60
vậy x=0 hoặc 60
\(x\left(x-5.12\right)=0\)
\(x\left(x-60\right)=0\)
\(x=0\) hoặc \(x-60=0\)
⇒ \(x=60\)
Vậy \(x\left\lbrace0;60\right\rbrace\)
Đặt \(A=\frac{25}{1\cdot6}+\frac{25}{6\cdot11}+\cdots+\frac{25}{25\cdot31}\)
\(A=5\cdot\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\cdots+\frac{5}{26\cdot31}\right)\)
\(A=5\cdot\left(\frac11-\frac16+\frac16-\frac{1}{11}+\cdots+\frac{1}{26}-\frac{1}{31}\right)\)
\(A=5\cdot\left(\frac11-\frac{1}{31}\right)\)
\(A=5\cdot\frac{30}{31}\)
\(A=\frac{150}{31}\)
Vậy \(A=\frac{150}{31}\)
\(\frac{25}{1*6}+\frac{25}{6*11}+...+\frac{25}{26*31}\)
=\(\frac{25}{1}-\frac{25}{6}+\frac{25}{6}-\frac{25}{11}+...+\frac{25}{26}-\frac{25}{31}\)
=\(\frac{25}{1}-\frac{25}{31}\)
=\(\frac{775}{31}-\frac{25}{31}\)
=\(\frac{750}{31}\)