tim x biet
\(1+3+5+7+.......+\left(2x+1\right)=5050\)
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tim x biet
\(3x\left(x-1\right)-x\left(3x-2\right)=5\)
\(8x\left(2x+1\right)-4x\left(2x-3\right)=-40\)
\(\left(2x-1\right)\left(3x-1\right)-\left(3x-2\right)\left(2x-1\right)=3\)
a ) \(3x\left(x-1\right)-x\left(3x-2\right)=5\)
\(\Leftrightarrow3x^2-3x-3x^2+2x=5\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy phương trình có nghiệm x = - 5 .
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a, \(3x\left(x-1\right)-x\left(3x-2\right)=5\)
\(\Rightarrow3x^2-3x-\left(3x^2-2x\right)=5\)
\(\Rightarrow3x^2-3x-3x^2+2x=5\)
\(\Rightarrow5x=5\Rightarrow x=1\)
Câu b,c làm tương tự! Cứ tách ra là làm được à!
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b ) \(8x\left(2x+1\right)-4x\left(2x-3\right)=-40\)
\(\Leftrightarrow16x^2+8x-8x^2+12x=-40\)
\(\Leftrightarrow20x=-40\)
\(\Leftrightarrow x=-2\)
Vậy phương trình có nghiệm x = - 2
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Xem thêm câu trả lời
1) Tinh :a) \(\frac{3^6.45^4-15^{13}.5^{-9}}{27^4.25^3+45^6}-\frac{50}{7}.\frac{\sqrt{0,49}}{25}\)
b) Tim x , biet: \(|2x-\frac{1}{5}|\left(-\frac{1}{5}\right)^5=\left(-\frac{1}{5}\right)^7\)
tim so huu ti x biet :
a)\(\left(2x-3\right)^4=\left(2x-3\right)^6\)
b) \(\left(3x+5\right)^3=\left(3x+5\right)^{2016}\)
c) \(\left(2x+1\right)^{2015}=\left(2x+1\right)^{2017}\)
a, (2x-3)4=(2x-3)6
=> (2x-3)6 : (2x-3)4=1
=> (2x-3)3=
=> 2x-3=1
=> 2x=4
=> x=2
b, (3x+5)3=(3x+5)2016
=> (3x+5)2016 : (3x+5)3=1
=> (3x+5)2013=1
=> 3x+5=1
=> 3x=-4
=> x=-4/3
c, (2x+1)2015=(2x+1)2017
=> (2x+1)2017 : (2x+1)2015=1
=> (2x+1)2=1
=> 2x+1=1
=> 2x=0
=> x=0
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Bai 1:a)Tim x biet\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\times\left(x+1\right)}=\frac{2009}{2011}\)
b)\(\left(x-1\right)\times f\left(x\right)=\left(x+4\right)\times f\left(x\right)\)voi moi x
Bai 2;Tim x;y;z biet a)\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}\) b)\(\frac{2x+1}{5}=\frac{3y-z}{7}=\frac{2x+3y-1}{6x}\)
3, tim x, biet ;
a,\(3x\left[12x-4\right]-9x\left[4x-3\right]=30\)
b,\(x\left[5-2x\right]+2x\left[x-1\right]=15\)
a) \(36x^2-12x-36x^2+27x=30\)
\(15x=30\)
\(x=2\)
b) \(5x-2x^2+2x^2-2x=15\)
\(3x=15\)
\(x=5\)
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1)Tim x, biet : a) \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
b) \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
4(18-5x)-12(3x-7)=15(2x-16)-6(x+14)
<=>72-20x-36x+84=30x-240x-6x-84
<=>160x=-86
<=>x=-0.0375
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tim x biet : \(\left|2x-5\right|+1=3x\)
Ta có:
|2x-5|+1=3x
=> |2x-5|=3x-1
=> 2x-5=3x-1 ; 2x-5=1-3x
=> -5+1=3x-2x ; 2x+3x=1+5
=>x=-4 ; 5x =6
; x=\(\dfrac{5}{6}\)
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Tim x biet
k) \(\left[\left(3,75:\frac{1}{4}+2\frac{2}{5}.125\%\right)-\left(\frac{7}{2}.0,8-1,2:\frac{3}{2}\right)\right]:\left(1\frac{1}{2}+0,75\right)x=64\)
Tim x biet
\(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
1/2(2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3))=15/93
1/2(1/3-1/5+1/5-1/7+1/7-1/9+...+1/2x+1-1/2x+3)=15/93
1/2(1/3-1/2x+3)=15/93
=>1/3-1/2x+3=10/31
=>1/2x+3=1/93
=>2x+3=93
2x=93-3=90
=>x=45
Đặt \(A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
\(\Rightarrow2A=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{3}-\frac{1}{2x+3}=\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}\)
\(\frac{1}{2x+3}=\frac{1}{93}\)
\(\Rightarrow2x+3=93\)
\(2x=90\)
\(x=45\)
Vậy \(x=45\).
\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)
=> \(\frac{1}{2}\left(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{5}{31}\)
=> \(\frac{1}{2}\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{5}{31}\)
=> \(\frac{1}{3}-\frac{1}{2x+3}=\frac{5}{31}:\frac{1}{2}=\frac{5}{31}\cdot2=\frac{10}{31}\)
=> \(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}=\frac{1}{93}\)
=> 2x +3 = 93
=> 2x = 90 => x = 45