giúp mình giải bài này nghen: 0,4+x-21/5-2,7=3,15-x+y-2,5-17/5+4,5-14,1=-0,7+z+t-44/5=-43/4-1,75-t-k=9,8+k-3,02+z .Tính x,y,z,t,k.
0,4+x-21/5-2.7=3,75-x+y-2,25=2,5-17/5+4,5-14,1=-0,7+z+t-44/5=-43/4-1,75-t-k=9,8+k-3,02+z
Tìm các số x, y, z, t, k, sao cho ta có đẳng thức sau:
\(0,4+x-4\frac{1}{5}-2,7=3,75-x+y-2,25=\)
\(=2,5-3\frac{2}{5}+4,5-14,1=-0,7+z+t-8\frac{4}{5}=\)
\(=-10\frac{3}{4}-1,75-t-k=9,80+k-3,02+z.\)
Tìm X; Y; Z ; T ; K biết :
0,4 + X + 4 1/5 - 2,7 = -3,75 - X + Y - 2,25 = 2,5 - 3 2/5 + 4,5 - 14,1 = 2,5 - 3 2/5 + 4,5 - 14,1 = -6,7 + 2 + T - 8 4/5 = -10 3/4 - 1,75 - T - K = 8,8 + K - 3,02 +Z
Giúp mk vs mọi người
mai mk phải nộp bài rùi giúp mk vs
Tìm các số x, y, z, t, k, sao cho ta có các đẳng thức sau:
\(0,4+x-4\frac{1}{5}-2,7=3,75-x+y-2,25=\)
\(=2,5-3\frac{2}{5}+4,5-14,1=-0,7+z+t-8\frac{4}{5}=\)
\(=-10\frac{3}{4}-1,75-t-k=9,80+k-3,02+z\)
Giải đầy đủ giúp mk nha
Tìm các số x,y,z,t,k, sao cho ta có đẳng thức sau:
\(0,4+x-4\frac{1}{5}-2,7=3,75-x+y-2,25=\)
\(=2,5-3\frac{2}{5}+4,5-14,1=-0,7+z+t-8\frac{4}{5}=\)
\(=-10\frac{3}{4}-1,75-t-k=9,80+k-3,02+z.\)
Bài 9: Tìm x, y, z, t nguyên biết
12/-6=x/5=-y/3=z/-17=-t/-9
Bài 10: Tìm x, y, z, t, u biết:
4/3=12/9=8/x=y/21=40/2=16/t=u/111
Bài 11: Tìm x, y, z, t, u biết:
-7/6=x/18=-98/y=-14/z=t/102=u/-78
Mong mn giải nhanh giúp ạ.
Mk đang cần gấp để nộp cho thầy ạ
Bài 9:
Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)
\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)
Vậy: (x,y,z,t)=(-10;6;34;-18)
Bài 11:
Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)
\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)
Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)
\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)
Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)
\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)
Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)
\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)
Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)
\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)
Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)
\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)
Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)
Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ
cho x,y,z,t,k là các số nguyên thỏa x+y+z+t+k chia hết cho 5
cmr \(x^5+y^5+z^5+t^5+k^5\)chia hết cho 5
tui chịu mấy má
Tìm k nếu x + y = 4, y + 5 =z, z + x =2k, x + y + z = 21
A.-12 B.-6 C.\(\dfrac{43}{6}\) D.7 E.\(\dfrac{51}{7}\)
Lời giải:
$z=(x+y+z)-(x+y)=21-4=17$
$y=z-5=17-5=12$
$2k=z+x=(x+y+z)-y=21-12=9$
$k=\frac{9}{2}$
Không đáp án nào đúng.
Câu 1 The function mm is defined on the real numbers by m(k) = \dfrac{k+2}{k+8}m(k)= k+8 k+2 . What is the value of 10\times m(2)10×m(2)? Answer: Câu 2 The function ff is defined on the real numbers by f(x)= ax-3f(x)=ax−3. What is the value of a if f(3)=9f(3)=9? Answer: Câu 3 The function ff is defined on the real numbers by f(x)= 2x+a-3f(x)=2x+a−3. What is the value of a if f(-5)=11f(−5)=11? Answer: Câu 4 The function ff is defined on the real numbers by f(x) = 2 + x-x^2f(x)=2+x−x 2 . What is the value of f(-3)f(−3)? Answer: Câu 5 Given a real number aa and a function ff is defined on the real numbers by f(x)=-6\times|3x|-4f(x)=−6×∣3x∣−4. Compare: f(a)f(a) f(-a)f(−a) Câu 6 There are ordered pairs (x;y)(x;y) where xx and yy are integers such that \dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8} x 5 + 4 y = 8 1 Câu 7 Given a negative number kk and a function ff is defined on the real numbers by f(x)=\dfrac{6}{13}xf(x)= 13 6 x. Compare: f(k)f(k) f(-k)f(−k) Câu 8 Given a positive number kk and a function ff is defined on the real numbers by f(x)=\dfrac{-3}{4}x+4f(x)= 4 −3 x+4. Compare: f(k)f(k) f(-k)f(−k). Câu 9 A=(1+2+3+\ldots+90) \times(12 \times34-6 \times 68):(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})=A=(1+2+3+…+90)×(12×34−6×68):( 3 1 + 4 1 + 5 1 + 6 1 )= Câu 10 Given that \dfrac{2x+y+z+t}{x}=\dfrac{x+2y+z+t}{y}=\dfrac{x+y+2z+t}{z}=\dfrac{x+y+z+2t}{t} x 2x+y+z+t = y x+2y+z+t = z x+y+2z+t = t x+y+z+2t . The negative value of \dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z} z+t x+y + t+x y+z + x+y z+t + y+z t+x is