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Ta có: \(\frac{t}{x}\)= \(\frac{4}{3}\)=\(\frac{8}{12}\) \(\frac{z}{x}\)=\(\frac{1}{6}\)=\(\frac{2}{12}\)
\(\frac{y}{z}\)=\(\frac{3}{2}\)
Suy ra: \(\frac{t}{y}\)=\(\frac{8}{3}\)
Bài1:
\(\dfrac{\left(1,09-0,29\right).\left(\dfrac{5}{4}\right)}{18,9-16,65.\left(\dfrac{8}{9}\right)}=\dfrac{\dfrac{4}{5}.\left(\dfrac{5}{4}\right)}{\left(\dfrac{9}{8}\right).\left(\dfrac{8}{9}\right)}=1\)
Bài 1:
\(A=\dfrac{\left(1,09-0,29\right)\cdot\dfrac{5}{4}}{\left(18,9-16,65\right)\cdot\dfrac{8}{9}}=\dfrac{0,8\cdot1,25}{2,25\cdot\dfrac{8}{9}}=\dfrac{1}{2}\)
\(B=\left[0,8\cdot7+\left(0,8\right)^2\right]\left(1,25\cdot7-\dfrac{4}{5}\cdot1,25\right)+31,64\)
\(=0,8\cdot\left(7+0,8\right)\cdot1,25\left(7-0,8\right)+31,64\)
\(=0,8\cdot7,8\cdot1,25\cdot6,2+31,64\)
\(=6,24\cdot7,75+31,64=48,36+31,65=80\)
\(\Rightarrow A:B=\dfrac{1}{2}:80=\dfrac{1}{160}\)
Vậy A gấp 1/160 lần B
bài 2:
\(\dfrac{x}{4}-\dfrac{1}{y}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{y}=\dfrac{x}{4}-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{y}=\dfrac{x-2}{4}\)
=>y(x-2)=4
=>y và x-2 thuộc Ư(4) = {1;-1;2;-2;4;-4}
Ta có bẳng:
| y | 1 | -1 | 2 | -2 | 4 | -4 |
| x-2 | 4 | -4 | 2 | -2 | 1 | -1 |
| x | 6 | -2 | 4 | 0 | 3 | 1 |
Vậy....
bài 3:
Ta có: x-y=x:y => x=xy+y=y(x+1) => x:y=y(x+1):y=x+1 (1)
Mà x:y=x-y (2)
Từ (1) và (2) => y = -1
Lại có: x=y(x+1) => x=(-1)(x+1) => x=-x-1 => 2x=-1 => x=\(\dfrac{-1}{2}\)
Vậy x=-1/2, y=-1
bài 4:
Ta có: x(x+y+z)+y(x+y+z)+z(x+y+z)=-5+9+5
=>(x+y+z)2=9
=>x+y+z=3 hoặc x+y+z=-3
Nếu x+y+z=3 => \(\left\{{}\begin{matrix}3x=-5\\3y=9\\3z=5\end{matrix}\right.\) =>\(\left\{{}\begin{matrix}x=\dfrac{-5}{3}\\y=3\\z=\dfrac{5}{3}\end{matrix}\right.\)
Nếu x+y+z=-3 => \(\left\{{}\begin{matrix}-3x=-5\\-3y=9\\-3z=5\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-3\\z=\dfrac{-5}{3}\end{matrix}\right.\)
Vậy....
\(\text{Câu 1: }\\ \text{Theo bài ra ta có : }x+y-z=10\\ \dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{2}=\dfrac{4y}{12}\Rightarrow\dfrac{x}{8}=\dfrac{y}{12}\left(1\right)\\ \dfrac{y}{4}=\dfrac{z}{5}\Rightarrow\dfrac{3y}{12}=\dfrac{z}{5}\Rightarrow\dfrac{y}{12}=\dfrac{z}{15}\left(2\right)\\ \text{Từ }\left(1\right)\text{ và }\left(2\right)\text{ suy ra : }\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}\\ \text{ Áp dụng tính chất dãy tỉ số bằng nhau ta được : }\\ \dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{8}=2\Rightarrow x=16\\\dfrac{y}{12}=2\Rightarrow y=24\\\dfrac{z}{15}=2\Rightarrow z=30\end{matrix}\right.\\ \text{Vậy }x=16\\ y=24\\ z=30\)
\(\text{Câu 2 : }\\ \text{Ta có : }\dfrac{x}{2}=\dfrac{y}{5}\\ \Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{5}\right)^2=\dfrac{x}{2}\cdot\dfrac{y}{5}=\dfrac{xy}{2\cdot5}=\dfrac{7+3}{10}=\dfrac{10}{10}=1\\ \Rightarrow\left\{{}\begin{matrix}\left(\dfrac{x}{2}\right)^2=1\Rightarrow\dfrac{x}{2}=1\Rightarrow x=2\\\left(\dfrac{y}{5}\right)^2=1\Rightarrow\dfrac{y}{5}=1\Rightarrow y=5\end{matrix}\right.\\ \text{Vậy }x=2\\ y=5\)
Câu 3 : \(\dfrac{\text{Giải}}{ }\)
Gọi số học sinh 4 khối \(6,7,8,9\) lần lượt là \(a;b;c;d\) \(\left(a;b;c;d\in N\text{*}\right)\) \(\left(em\right)\)
Theo bài ra ta có : \(b-d=70\)
\(a;b;c;d\) tỉ lệ với \(9;8;7;6\) \(\Rightarrow\dfrac{a}{9}=\dfrac{b}{8}=\dfrac{c}{7}=\dfrac{d}{6}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được :
\(\dfrac{a}{9}=\dfrac{b}{8}=\dfrac{c}{7}=\dfrac{d}{6}=\dfrac{b-d}{8-6}=\dfrac{70}{2}=35\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{9}=35\Rightarrow a=315\\\dfrac{b}{8}=35\Rightarrow b=280\\\dfrac{c}{7}=35\Rightarrow c=245\\\dfrac{d}{6}=35\Rightarrow d=210\end{matrix}\right.\)
\(\text{Vậy }a=315\\ b=280\\ c=245\\ d=210\)
1, ta co \(\frac{x}{5}=\frac{y}{6}=\frac{x}{20}=\frac{y}{24}\)
\(\frac{y}{8}=\frac{z}{7}=\frac{y}{24}=\frac{z}{21}\)
=>\(\frac{x}{20}=\frac{y}{24}=\frac{z}{21}=\frac{x+y-z}{20+24-21}=\frac{69}{23}=3\)
=>\(x=3\cdot20=60\)
\(y=3\cdot24=72\)
\(z=3\cdot21=63\)
3. ta co \(\frac{x}{15}=\frac{y}{7}=\frac{z}{3}=\frac{t}{1}=\frac{x+y-z+t}{15-7+3-1}=\frac{10}{10}=1\)
=> \(x=1\cdot15=15\)
\(y=1\cdot7=7\)
\(z=1\cdot3=3\)
\(t=1\cdot1=1\)
bài này hay, áp dụng t/c tỷ lệ thức có;
=\(\frac{y+x+y+x}{x-z+z+y}\)= 2(x+y)/(x+y) =2
<=> x/y = 2
\(\frac{z}{x}=\frac{1}{6}\Rightarrow\frac{x}{z}=6\)
\(\Rightarrow\frac{t}{x}.\frac{x}{z}=\frac{t}{z}=8\)
\(\frac{y}{z}=\frac{3}{2}\Rightarrow\frac{z}{y}=\frac{2}{3}\)
\(\Rightarrow\frac{t}{z}.\frac{z}{y}=\frac{t}{y}=\frac{16}{3}\)
1 a) \(\dfrac{\left(-2\right)}{5}\)= \(\dfrac{-6}{15}\); \(\dfrac{15}{-6}\)= \(\dfrac{5}{-2}\); \(\dfrac{-6}{-2}\)= \(\dfrac{15}{5}\); \(\dfrac{-2}{-6}\)= \(\dfrac{5}{15}\)
Bài 1 :
a/ \(x^2-7x+6=0\)
\(\Leftrightarrow x^2-6x-x+6=0\)
\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=1\end{matrix}\right.\)
Vậy....
b/ \(x^2-10x+9=0\)
\(\Leftrightarrow x^2-9x-x+9=0\)
\(\Leftrightarrow x\left(x-9\right)-\left(x-9\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-9=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy...
c/ \(x^2+9x+8=0\)
\(\Leftrightarrow x^2+8x+x+8=0\)
\(\Leftrightarrow\left(x+8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x+1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-1\end{matrix}\right.\)
Vậy ...
d/ \(x^2-11x+10=0\)
\(\Leftrightarrow x^2-11x+10=0\)
\(\Leftrightarrow x^2-x-10x+10=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=10\end{matrix}\right.\)
Vậy...
Bài 2 :
Ta có :
\(\frac{2x-y}{x+y}=\frac{2}{3}\)
\(\Leftrightarrow3\left(2x-y\right)=2\left(x+y\right)\)
\(\Leftrightarrow6x-3y=2x+2y\)
\(\Leftrightarrow6x-2x=2y+3y\)
\(\Leftrightarrow4x=5y\)
\(\Leftrightarrow\frac{x}{y}=\frac{5}{4}\)
Vậy....
Bài 3 : không hiểu đề lắm ???!!!!
Bài 4 :
Ta có :
\(\frac{x}{y^2}=2\Leftrightarrow x=2y^2\left(1\right)\)
Thay (1) ta có :
\(\frac{x}{y}=16\)
\(\Leftrightarrow\frac{2y^2}{y}=16\)
\(\Leftrightarrow2y=16\)
\(\Leftrightarrow y=8\Leftrightarrow x=128\)
Vậy...
a)