Tìm x biết
2x/15+2x/35+2x/63+....+2x/195=4/5
Chứng minh rằng 1/2-1/4+1/8-1/16+1/32-1/1/64<1/3
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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
17x + 3. ( -16x – 37) = 2x + 43 - 4x
<=>17x-48x-111=-2x+43
<=>-29x=154
<=> \(x=-\frac{154}{29}\)
-3. (2x + 5) -16 < -4. (3 – 2x)
\(\Leftrightarrow-6x-31< -12+8x.\)
\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)
Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu
a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14)
=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84
=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84)
=> 156 - 56x = 24x - 324
=> 24x + 56x = 324 + 156
=> 80x = 480
=> x = 480 : 80 = 6
Vậy x = 6
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
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
\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)
Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)
Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)