tìm x:
4/5x3/7xX/3=24/105
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=>5x^3+4x^2+3x+3-4+x+4x^2-5x^3=5
=>8x^2+4x-1-5=0
=>8x^2+4x-6=0
=>4x^2+2x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{4}\)
\(\left(x.7-7\right)\left(x.12+24\right)=0\)
=> \(\orbr{\begin{cases}x.7-7=0\\x.12+24=0\end{cases}}\)
=> \(\orbr{\begin{cases}7x=7\\12x=-24\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
\(170-\left(4+5x\right)=101\)
=> \(4+5x=170-101\)
=> \(4+5x=69\)
=> \(5x=69-4\)
=> \(5x=65\)
=> \(x=65:5=13\)
(x.7-7).(x.12+24)=0
* x.7-7=0 * x.12+24=0
x.7=7 x.12=24
x=7:7 x=24:12
x=1 x=2
vạy x=1 hoặc x=2
170-(4+5x)=101
4+5x=170-101
4+5x=69
5x=69-4
5x=65
x=65:5
x=13
vậy x=13
250:(2x-4)=145-120
250:(2x+4)=25
2x+4=250:25
2x+4=10
2x=10+4
2x=14
x=14:2
x=7
vậy x=7
6.x+3.x-4.x=105
x.(6+3-4)=105
x.(9-4)=105
x.5=105
x=105:5
x=21
vậy x=21
\(2016\times105-2016\times4-2016\)
\(=2016\times\left(105-4-1\right)\)
\(=2016\times100\)
\(=201600\)
\(x\times24+x\times6=240\)
\(x\times\left(24+6\right)=240\)
\(x\times30=240\)
\(x=240:30\)
\(x=8\)
\(x-167\times15=167\times185\)
\(x-2505=30895\)
\(x=30895+2505\)
\(x=33400\)
\(173\times105+173\times96-173\)
\(=173\times\left(105+96-1\right)\)
\(=173\times200\)
\(=34600\)
1. 2016 x 105 - 2016 x 4 - 2016
= 211680 - 8064 - 2016
= 203616 - 2016
= 201600
2. X x 24 + X x 6 = 240
X x (24 + 6) = 240
X x 30 = 240
X = 240 : 30
X = 8
3. X - 167 x 15 = 167 x 185
X - 2505 = 30895
X = 30895 - 2505
X = 28390
4. 173 x 105 + 173 x 96 - 173
= 18165 + 16608 - 173
= 34773 - 173
= 34600
f ) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Đặt \(x^2+5x+5=t\), ta có :
\(\left(t-1\right)\left(t+1\right)-24\)
\(=t^2-1-24=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)\)
Thay và ta có :
\(\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(=x\left(x+5\right)\left(x^2+5x+10\right)\)
a) \(58+7x=100\)
\(=>7x=100-58\)
\(=>7x=42\)
\(=>x=42:7\)
\(=>x=6\)
b) \(3x-7=28\)
\(=>3x=28+7\)
\(=>3x=35\)
\(=>x=35:3\)
\(=>x=\dfrac{35}{3}\)
c) \(x-56:4=16\)
\(=>x-14=16\)
\(=>x=16+14\)
\(=>x=30\)
d) \(101+\left(36-4x\right)=105\)
\(=>36-4x=105-101\)
\(=>36-4x=4\)
\(=>4x=36-4\)
\(=>4x=32\)
\(=>x=32:4\)
\(=>x=8\)
e) \(\left(x-12\right):12=12\)
\(=>x-12=12.12\)
\(=>x-12=144\)
\(=>x=144-12\)
\(=>x=132\)
f) \(\left(3x-2^4\right).7^3=2.7^4\)
\(=>3x-2^4=2.7^4:7^3\)
\(=>3x-16=2.7=14\)
\(=>3x=14+16\)
\(=>3x=30\)
\(=>x=30:3\)
\(=>x=10\)
i) \(\left(10+2x\right).4^{2011}=4^{2013}\)
\(=>10+2x=4^{2013}:4^{2011}\)
\(=>10+2x=4^2=16\)
\(=>2x=16-10\)
\(=>2x=6\)
\(=>x=6:2\)
\(=>x=3\)
\(#WendyDang\)
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
1) Ta có: \(5\left(x-3\right)\left(x-7\right)-\left(5x+1\right)\left(x-2\right)=-8\)
\(\Leftrightarrow5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=-8\)
\(\Leftrightarrow5x^2-50x+105-5x^2+9x+2+8=0\)
\(\Leftrightarrow-41x=-115\)
hay \(x=\dfrac{115}{41}\)
2) Ta có: \(x\left(x+1\right)\left(x+2\right)-\left(x+4\right)\left(3x-5\right)=84-5x\)
\(\Leftrightarrow x\left(x^2+3x+2\right)-\left(3x^2+7x-20\right)=84-5x\)
\(\Leftrightarrow x^3+3x^2+2x-3x^2-7x+20-84+5x=0\)
\(\Leftrightarrow x^3=64\)
hay x=4
3) Ta có: \(\left(9x^2-5\right)\left(x+3\right)-3x^2\left(3x+9\right)=\left(x-5\right)\left(x+4\right)-x\left(x-11\right)\)
\(\Leftrightarrow9x^3+27x^2-5x-15-9x^3-27x^2=x^2-x-20-x^2+11x\)
\(\Leftrightarrow-5x-15=10x-20\)
\(\Leftrightarrow-5x-10x=-20+15\)
\(\Leftrightarrow x=\dfrac{-5}{-15}=\dfrac{1}{3}\)
a, 100 - 7 x ( x - 5 ) = 58
7 x ( x - 5 ) = 100 - 58
7 x ( x - 5 ) = 42
( x - 5 ) = 42 : 7
x - 5 = 6
x = 6 + 5
x = 11
b, 24 + 5 * x = 49
5 * x = 49 - 24
5 * x = 25
x = 25 : 5
x = 5
c, x - 105 : 21 = 15
x - 5 = 15
x = 15 + 5
x = 20
d, ( x - 105 ) : 21 = 15
( x - 105 ) = 15 x 21
x - 105 = 315
x = 315 + 105
x = 420
\(\frac{4}{5}\)\(\times\)\(\frac{3}{7}\)\(\times\)\(\frac{x}{3}\)=\(\frac{24}{105}\)
= \(\frac{4\times3}{5\times7}\)\(\times\)\(\frac{x}{3}=\frac{24}{105}\)
=\(\frac{12}{35}\times\frac{x}{3}=\frac{24}{105}\)
\(\Rightarrow\frac{x}{3}=\frac{24}{105}:\frac{12}{35}\)
=\(\frac{x}{3}=\frac{2}{3}\)
Vậy X = 2
4/5 x 3/7 x X/3 = 24/105
12/35 x X/3 = 24/105
X/3 = 24/105 : 12/35
X/3 = 2/3
X = 2/3 x 3
X = 2