1. Tìm x,y biết : \(\dfrac{3}{5}x\) =\(\dfrac{2}{3}y\) và x2-y2= 38
2. Cho:\(\dfrac{x +16}{19}\)=\(\dfrac{y-25}{16}\)=\(\dfrac{z+9}{25}\) và 2x3-1=15. tính x+y+z
1. Tìm x,y biết : \(\dfrac{3}{5}x\) =\(\dfrac{2}{3}y\) và x2-y2= 38
2. Cho:\(\dfrac{x +16}{19}\)=\(\dfrac{y-25}{16}\)=\(\dfrac{z+9}{25}\) và 2x3-1=15. tính x+y+z
2. Tham khảo thêm tại đây nha bạn
https://hoc24.vn/hoi-dap/question/417550.html
Tính
\(\left(-0,75.\dfrac{-1}{4}\right):\left(-5\right)+\dfrac{1}{5}+\dfrac{1}{5}-\left(\dfrac{-1}{5}\right):\left(-3\right)\)
\(\left(-0,75.\dfrac{-1}{4}\right):\left(-5\right)+\dfrac{1}{5}+\dfrac{1}{5}-\left(\dfrac{-1}{5}\right):\left(-3\right)\)
\(=\left(\dfrac{-3}{4}.\dfrac{-1}{4}\right):\left(-5\right)+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}:\left(-3\right)\)
\(=\left(\dfrac{3}{16}\right):\left(-5\right)+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{1}{5}.\dfrac{-1}{3}\)
\(=\dfrac{-3}{80}+\dfrac{1}{5}+\dfrac{1}{5}+\dfrac{-1}{15}\)
\(=\dfrac{71}{240}\)
Tìm GTNN của biểu thức sau:
a) A = (6x-1)^2 + 2017
b) B = (x^2-16)^4 + (y-2)^2
c) C = 15 + I 2x-1 I ( I là dấu trị tuyệt đối )
d) D = (x-1)^2 + (2x-y)^2 + 3
a) \(A=\left(6x-1\right)^2+2017\)
Vì \(\left(6x-1\right)^2\ge0\)
Nên \(\left(6x-1\right)^2+2017\ge2017\)
Vậy GTNN của A=2017 khi \(6x-1=0\Leftrightarrow x=\dfrac{1}{6}\)
c) \(C=15+\left|2x-1\right|\)
Vì \(\left|2x-1\right|\ge0\)
Nên \(\left|2x-1\right|+15\ge15\)
Vậy GTNN của C=15 khi \(2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)
d) \(D=\left(x-1\right)^2+\left(2x-y\right)^2+3\)
Vì \(\left(x-1\right)^2+\left(2x-y\right)^2\ge0\)
Nên \(\left(x-1\right)^2+\left(2x-y\right)^2+3\ge3\)
Vậy GTNN của D=3 khi \(\left\{{}\begin{matrix}x-1=0\\2x-y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=1\\2.1-y=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)
Có bn nào đã ktra 1 tiết toán số 7 chưa???
có thì gửi đề cho mk nha>>>mk tick cho
Chương I
Tìm n:
a) \(\left(\dfrac{1}{2}\right)^n=\dfrac{1}{32}\)
b) \(\left(\dfrac{343}{125}\right)=\left(\dfrac{7}{5}\right)^n\)
c) \(\dfrac{16}{2^n}=2\)
d) \(\dfrac{\left(-3\right)^n}{81}=-27\)
e) 8n : 2n =4
f) 32 . 3n = 35
g) ( 22 : 4 ).2n = 4
h) 3-2 . 34 . 3n = 37
a)\(\left(\dfrac{1}{2}\right)^n=\dfrac{1}{32}\)
=>\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^5\)
=>n=5
b)\(\left(\dfrac{343}{125}\right)=\left(\dfrac{7}{5}\right)^n\)
=>\(\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\)
=>n=3
c)\(\dfrac{16}{2^n}=2\)
=>2n=\(\dfrac{16}{2}\)
=>2n=8
=>2n=23
=>n=3
d)\(\dfrac{\left(-3\right)^n}{81}=-27\)
=>(-3)n=-27.81
=>(-3)n=-2187
=>(-3)n=(-3)7
=>n=7
e)8n:2n=4
=>(23)n:2n=4
=>23n:2n=4
=>23n-n=4
=>22n=4
=>22n=22
=>2n=2
=>n=1
f)32.3n=35
=>3n=35:32
=>3n=35-2
=>3n=33
=>n=3
g) (22:4).2n=4
=>1.2n=22
=>n=2
h)3-2.34.3n=37
=>\(\left(\dfrac{1}{3}\right)^2\).34.3n=37
=>32.3n=37
=>32+n=37
=>2+n=7
=>n=5
1) Tìm x,y,z biết
a) x.y=2/3; yz=0.6; xz0.625
b) (x+2)^2+(y-3)^4+(z-5)^6=0
c) x(x-y+z)=-11; y(y-z-z)=25 và z(z+x-y)=35
2) Tìm x biết
a) x-1/65+x-3/63=x-5/61+x-7/59
Bài 2:
Ta có: \(\dfrac{x-1}{65}+\dfrac{x-3}{63}=\dfrac{x-5}{61}+\dfrac{x-7}{59}\)
\(\Leftrightarrow\left(\dfrac{x-1}{65}-1\right)+\left(\dfrac{x-3}{63}-1\right)=\left(\dfrac{x-5}{61}-1\right)+\left(\dfrac{x-7}{59}-1\right)\)
\(\Leftrightarrow\left(x-66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\right)=0\)
=>x-66=0
hay x=66
xác định a, để da thức \(f\left(x\right)=x^3+2x^2+ax+b\)chia hết cho đa thức \(g\left(x\right)=x^2+x+1\)
Chứng minh rằng :
\(A=75.\left(4^{1999}+a^{1998}+...+4^2+4+1\right)+25\) là số chia hết cho 100
Đặt \(B=1+4+4^2+...+4^{1998}+4^{1999}\)
\(\Rightarrow4B=4+4^2+4^3+...+4^{1999}+4^{2000}\)
\(\Rightarrow4B-B=\left(4+4^2+4^3+...+4^{2000}\right)-\left(1+4+4^2+...+4^{1999}\right)\)
\(\Rightarrow3B=4^{2000}-1\)
\(\Rightarrow B=\dfrac{4^{2000}-1}{3}\)
Khi đó ta có:
\(A=75.B=75.\dfrac{4^{2000}-1}{3}=\dfrac{75.\left(4^{2000}-1\right)}{3}=\dfrac{75}{3}.\left(4^{2000}-1\right)=25.\left(4^{2000}-1\right)=25.4^{2000}-25\)
Ta có: \(4^{2000}-1=\left(4^4\right)^{500}-1=\left(...6\right)-1=...5\)
\(\Rightarrow25.4^{2000}-25=25.\left(...5\right)-25=\left(...5\right)-25=...0⋮100\left(đpcm\right)\)
Ta có:
\(A=75.\left(4^{1999}+4^{1998}+...+4^2+4+1\right)+25\)
\(A=25.3.\left(4^{1999}+4^{1998}+...+4^2+4+1\right)+25\) \(A=25.\left(4-1\right).\left(4^{1999}+4^{1998}+...+4^2+4+1\right)+25\)
\(A=25.\left(4^{2000}+4^{1999}+...+4^3+4^2+4-4^{1999}-4^{1998}-...-4^2-4-1\right)+25\)\(A=25.\left(4^{2000}-1\right)+25\)
\(A=25.\left(4^{2000}-1+1\right)\)
\(A=25.4^{2000}=25.4.4^{1999}=100.4^{1999}\)Vây:A là số chia hết cho 100
So sánh : a) (-39)9 và (-18)13
b)\(\dfrac{1}{101^2}+\dfrac{1}{102^2}+\dfrac{1}{103^2}+\dfrac{1}{104^2}+\dfrac{1}{105^2}\) và \(\dfrac{1}{2^2.3.5^2.7}\)
Tính:
(1000 - 13).(1000 - 23 ).(1000 - 33 )......(1000 - 253)
\(\left(1000-1^3\right)\left(1000-2^3\right).............\left(1000-25^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)......\left(1000-25^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)......\left(1000-1000\right)......\left(1000-25^3\right)\)
\(=\left(1000-1^3\right)\left(1000-2^3\right)........0.......\left(1000-25^3\right)\)
\(=0\)