Tìm x:
(2x - 1) . (2x + 1) = 35
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a: =>31-x=60
=>x=-29
b: =>(x-140):35=280-270=10
=>x-140=350
=>x=490
c: =>(1900-2x):35=48
=>1900-2x=1680
=>2x=220
=>x=110
d: =>\(2^{2x-1}=2^9\cdot2=2^{11}\)
=>2x-1=11
=>x=6
e: =>(x+2)^5=4^5
=>x+2=4
=>x=2
f: =>3x-4=0 hoặc x-1=0
=>x=4/3 hoặc x=1
g: =>(2x-1)^2=49
=>2x-1=7 hoặc 2x-1=-7
=>x=-3 hoặc x=4
h: =>x(x+1)/2=78
=>x(x+1)=156
=>x=12
a.
$xy=-21=7.(-3)=(-7).3=3.(-7)=(-3).7=21.(-1)=(-21).1=(-1).21=1(-21)$
Do đó $(x,y)=(7,-3); (-7,3); (3,-7); (-3,7); (21,-1); (-21,1); (-1,21); (1,-21)$
b.
$(x+5)(y-3)=14=1.14=14.1=(-14)(-1)=(-1)(-14)=2.7=7.2=(-2)(-7)=(-7)(-2)$
Do đó:
$(x+5,y-3)=(1,14); (14,1); (-14,-1); (-1,-14); (2,7); (7,2); (-2,-7); (-7,-2)$
Đến đây thì đơn giản rồi.
c.
$x(y-2)=-19$, bạn làm tương tự
d. Tương tự
Bài giải chi tiết đây em nhé:
\(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\)+...+ \(\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\) = \(\dfrac{9}{19}\)
\(\dfrac{1}{2}\)(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+ \(\dfrac{2}{7.9}\)+...+ \(\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\)) = \(\dfrac{9}{19}\)
\(\dfrac{1}{2}\)( \(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)+ \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) +... + \(\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)
\(\dfrac{1}{2}\) ( 1 - \(\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)
1 - \(\dfrac{1}{2x+1}\) = \(\dfrac{9}{19}\) : \(\dfrac{1}{2}\)
1 - \(\dfrac{1}{2x+1}\) = \(\dfrac{18}{19}\)
\(\dfrac{1}{2x+1}\) = \(1-\dfrac{18}{19}\)
\(\dfrac{1}{2x+1}\) = \(\dfrac{1}{19}\)
\(2x+1\) = 19
2\(x\) = 19 - 1
2\(x\) = 18
\(x\) = 18: 2
\(x\) = 9
a: \(280-\left(x-140\right):35=270\)
=>(x-140):35=280-270=10
=>x-140=350
=>x=350+140
=>x=490
b: \(\left(190-2x\right):35-32=16\)
=>\(\left(190-2x\right):35=32+16=48\)
=>\(190-2x=35\cdot48=1680\)
=>2x=190-1680=-1490
=>x=-745
c: \(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=2^3\cdot5\)
=>\(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=8\cdot5=40\)
=>41-(2x-5)=720:40=18
=>2x-5=41-18=23
=>2x=28
=>x=14
d: \(\left(x:23+45\right)\cdot37-22=2^4\cdot105\)
=>\(\left(\frac{x}{23}+45\right)\cdot37=16\cdot105+22=1702\)
=>\(\frac{x}{23}+45=46\)
=>\(\frac{x}{23}=1\)
=>x=23
e: \(\left(3x-4\right)\left(x-1\right)^3=0\)
=>\(\left[\begin{array}{l}3x-4=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=1\end{array}\right.\)
f: \(2^{2x-1}:4=8^3\)
=>\(2^{2x-1-2}=2^9\)
=>2x-3=9
=>2x=12
=>x=6
g: \(x^{17}=x\)
=>\(x^{17}-x=0\)
=>\(x\left(x^{16}-1\right)=0\)
=>\(\left[\begin{array}{l}x=0\\ x^{16}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{16}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)
h: \(\left(x-5\right)^4=\left(x-5\right)^6\)
=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)
=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)
=>\(\left(x-5\right)^4\cdot\left(x-4\right)\left(x-6\right)=0\)
=>x∈{4;5;6}
i: \(\left(x+2\right)^5=2^{10}\)
=>\(\left(x+2\right)^5=\left(2^2\right)^5=4^5\)
=>x+2=4
=>x=2
k: 1+2+3+...+x=78
=>\(\frac{x\left(x+1\right)}{2}=78\)
=>x(x+1)=156
=>\(x^2+x-156=0\)
=>(x+13)(x-12)=0
=>x=-13(loại) hoặc x=12(nhận)
l: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)
=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)
=>3x=16+14=30
=>\(x=\frac{30}{3}=10\)
n: \(5^{x}:5^2=125\)
=>\(5^{x-2}=5^3\)
=>x-2=3
=>x=5
m: \(\left(x+1\right)^2=\left(x+1\right)^0\)
=>\(\left(x+1\right)^2=1\)
=>\(\left[\begin{array}{l}x+1=1\\ x+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-2\end{array}\right.\)
o: Số số hạng của dãy số 2;4;..;52 là:
(52-2):2+1=50:2+1=25+1=26(số)
Tổng của dãy số 2;4;...;52 là:
\(\left(52+2\right)\cdot\frac{26}{2}=54\cdot13=702\)
(2+x)+(4+x)+...+(52+x)=780
=>26x+702=780
=>26x=78
=>x=3
p: \(70=2\cdot5\cdot7;80=2^4\cdot5\)
=>ƯCLN(70;80)=\(2\cdot5=10\)
70⋮x; 80⋮x
=>x∈ƯC(70;80)
=>x∈Ư(10)
mà x>8
nên x=10
q: \(12=2^2\cdot3;25=5^2;30=2\cdot3\cdot5\)
=>BCNN(12;25;30)=\(2^2\cdot3\cdot5^2=300\)
x⋮12; x⋮25; x⋮30
=>x∈BC(12;25;30)
=>x∈B(300)
mà 0<x<500
nên x=300
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
4x^2 - 1= 35
x= +-3
Tìm x :( Bài này ngoài tầm lớp 5 nha )
\(\left(2x-1\right).\left(2x+1\right)=35\)
\(x=+_-3\)
nhé mí bạn