Tìm x
(x\(\times\)5)+(x\(\times\)2)+(x\(\times\)3)=100
Tìm x
A, ( x - 3 ) ^2 = (-3)^2 + (-4)^2
B, -12 × ( x -5 ) + 7 × ( 3-x) = 5
C, 30 × ( x + 2) -6 × ( x-5 ) - 24 × x = 100
a) \(\left(x-3\right)^2\) = \(\left(-3\right)^2\) + \(\left(-4\right)^2\)
\(\left(x-3\right)^2=9+16\)
\(\left(x-3\right)^2=25\)
\(\left(x-3\right)^2=5^2\)
\(\Rightarrow x-3=5\)
\(\Rightarrow x=5+3=8\)
Vậy x = 8
b) -12 . (x - 5) +7 . (3 - x) = 5
-12x + 60 + 21 - 7x = 5
-12x - 7x = 5 - 60 - 21
-19x = -76
x = -76 : (-19) =4
Vậy x = 4
c) 30 . (x + 2) - 6 . ( x - 5) - 24x = 100
30x + 60 - 6x + 30 - 24x =100
30x - 6x - 24x = 100 - 30 - 60
0x = 10
x=10 : 0
\(\Rightarrow x\in\varnothing\)
Tìm x , biết
a, -12 × ( x - 5 ) + 7 × ( 3 × x ) = 5
b, 30 × ( x + 2 ) - 6 × ( x - 5 ) - 24 × x = 100
c, | 5 × x - 2 | < 13 ( x thuộc Z )
a) -12.(x-5)+7.(3.x)=5
<=> -12x+60+21+7x=5
<=>-5x+81=5
<=>-5x=5-81=-76
<=>x=-76/-5=76/5=15,2
b) 30.(x+2)-6.(x-5)-24.x=100
<=> 30x+60-6x+30-24x=100
<=> 0x=100-60-30=10
=> không có giá trị nào của x để 0x=10
c) \(|5.x-2|< 13\)
Khi 5x-2 < 13
<=> 5x<15 <=> x<3
Khi 5x-2 <-13
<=> 5x<-11 <=> x<-11/5 <=> x<-2,2
\(x\times x^2\times x^3\times x^4\times...\times x^{100}\)
\(x\times x^2\times x^3\times x^4\times...\times x^{100}=x^{1+2+3+4+...+100}=x^{101\times500}=x^{5050}\)
\(x\times x^2\times x^3\times x^4\times...\times x^{100}\)
\(=x^{\left(1+100\right)\times100\div2}\)
\(=x^{5050}\)
nha
\(x\times x^2\times x^3\times...\times x^{100}\)
Áp dụng công thức: Nhân 2 lũy thừa cùng cơ số.
Ta có:
\(x\times x^2\times x^3\times...\times x^{100}\)
\(=x^{1+2+3+...+100}\)
\(x=5050\)
\(x.x^2.x^3...x^{100}=x^{1+2+3+...+100}\)
Đặt \(3^{1+2+3+...+100}=3^A\)
Ta có:
\(A=1+2+3+...+100\)
\(\Rightarrow A=100+99+98+...+1\)
\(\Rightarrow A=\left(1+100\right)+\left(2+99\right)+\left(3+98\right)+...+\left(100+1\right)\) ( 50 cặp số )
\(\Rightarrow A=101+101+101+...+101\) ( 50 số 101 )
\(\Rightarrow A=101.50\)
\(\Rightarrow A=5050\)
\(\Rightarrow3^A=3^{5050}\)
Vậy \(x.x^2.x^3...x^{100}=x^{5050}\)
Dấu chấm thay cho dấu nhân nhé!
1.viết tích dưới dạng lũy thừa
a.5x×5x×5x
b.x mũ 1 × x mũ 2.......x mũ 2006
c.x mũ 1×x mũ 4 ×x mũ 7.......x mũ 100
d.x mũ 2 ×x mũ 5×x mũ 8.......x mũ 2003
2.tìm x,y
2 mũ x +80=3 mũ y
1.
a) \(5x.5x.5x=\left(5x\right)^3.\)
b) \(x^1.x^2.....x^{2006}=x^{\frac{\left(2006+1\right).2006}{2}=}x^{2013021}.\)
c) \(x^1.x^4.x^7.....x^{100}=x^{\frac{\left(100+1\right).\left(\frac{100-1}{3}+1\right)}{2}}=x^{1717}.\)
d) \(x^2.x^5.x^8.....x^{2003}=x^{\frac{\left(2003+2\right).\left(\frac{2003-2}{3}+1\right)}{2}}=x^{669670}.\)
2.
\(2^x+80=3^y\)
Với \(x>0\Rightarrow2^x\) chẵn
Và 80 chẵn
\(\Rightarrow2^x+80\) chẵn.
Mà \(3^y\) lẻ
\(\Rightarrow x< 0.\)
Mà \(x\in N\)
\(\Rightarrow x=0.\)
\(\Rightarrow2^0+80=3^y\)
\(\Rightarrow1+80=3^y\)
\(\Rightarrow3^y=81\)
\(\Rightarrow3^y=3^4\)
\(\Rightarrow y=4.\)
Vậy \(\left(x;y\right)=\left(0;4\right).\)
Chúc bạn học tốt!
1.viết tích dưới dạng lũy thừa
a.5x.5x.5x
=(5x)\(^3\)
b.x\(^1\) . x \(^2\).......x \(^{2006}\)
=x \(^{2013021}\)
c.x\(^1\).x \(^4\) .x \(^7\)......x \(^{100}\)
=x \(^{1717}\)
d.x \(^2\) .x \(^5\).x \(^8\).......x\(^{2003}\)
=x \(^{669670}\)
2 \(^x\)+80=3\(^y\)
ta có 3\(^y\)lẻ nên 2 \(^x\)+80 lẻ .
Với x \(\ge\)1
thì 2 \(^x\)+80 chẵn không thỏa mãn
nên x=0
=>1+80=3\(^y\)
=>81=3\(^y\)
=>3\(^4\)=3\(^y\)
=>y=4
Tìm x;y;z biết:
\(x:y:z=3:4:5\) và \(2\times x^2+2\times y^2-3\times z^2=-100\)
Giúp mình với mình cần gấp!
Ta có: x/3=y/4=z/5.......
2*x^2/2*3^2+2*y^2/2*4^2-3*z^2=-100/-25=4
x/3=4 suy ra x=12
y/4=4 ....y=16
z/5.......z=20
Ta co : x:y:z=3:4:5
Hay : x/3=y/4=z/5
=>2x^2/18=2y^2/32=3z^2/75 và 2x^2+2y^2-3z^2=-100
Áp dụng tính chất dãy tỉ số bằng nhau :
2x^2/18=2y^2/32=3z^2/75=2x^2+2y^2-3z^2/18+32-75=-100/-25=4
Suy ra : 2x^2/18=4=>2x^2=72=>x^2=36=>x=+6
2y^2/32=4=>2y^2=128=>y^2=64=>y=+8
3z^2/75=4=>3z^2=300=>z^2=100=>z=+10
k nha , k hiu ns mk
4. Tìm x,y bt
a, (x^2-1)×(x^2-4)×(x^2-7)×(x^2-10)<0
b, (x^3 +5 )×(x^3+10)×(x^3+15)×(x^3+20)<0
Tìm x biết
a, | 1/2 × x - 3 | = 1/2
b, | 3× x + 1/5 | - 1/3 = 2/3
c, |3× x - 1/5 | + | 15× x - 1 |= 6/2
d, | 1/2 × x - 3 | < 1/2
e, | 3× x - 4/5 | > 1/3
a: \(\left|\dfrac{1}{2}x-3\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-3=\dfrac{1}{2}\\\dfrac{1}{2}x-3=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\dfrac{7}{2}\\\dfrac{1}{2}x=\dfrac{5}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)
b: \(\left|3x+\dfrac{1}{5}\right|-\dfrac{1}{3}=\dfrac{2}{3}\)
\(\Leftrightarrow\left|3x+\dfrac{1}{5}\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{1}{5}=1\\3x+\dfrac{1}{5}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{4}{5}\\3x=-\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{15}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: \(\left|\dfrac{1}{2}x-3\right|< \dfrac{1}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-3>-\dfrac{1}{2}\\\dfrac{1}{2}x-3< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x>\dfrac{5}{2}\\\dfrac{1}{2}x< \dfrac{7}{2}\end{matrix}\right.\Leftrightarrow5< x< 7\)
e: \(\left|3x-\dfrac{4}{5}\right|>\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{4}{5}>\dfrac{1}{3}\\3x-\dfrac{4}{5}< -\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x>\dfrac{17}{15}\\3x< \dfrac{7}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{17}{45}\\x< \dfrac{7}{45}\end{matrix}\right.\)