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Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
a)
ĐKXĐ: \(2x\geq 0\Leftrightarrow x\geq 0\)
Vậy TXĐ của $x$ là \(D= [0;+\infty)\)
b)
ĐK: \((2x-1)(x+3)\neq 0\Leftrightarrow \left\{\begin{matrix} 2x-1\neq 0\\ x+3\neq 0\end{matrix}\right.\Leftrightarrow \Leftrightarrow \left\{\begin{matrix} x\neq \frac{1}{2}\\ x\neq -3\end{matrix}\right.\)
Vậy TXĐ \(D=\mathbb{R}\setminus \left\{\frac{1}{2}; -3\right\}\)
c)
ĐK: \(8x^3+1\neq 0\Leftrightarrow x^3\neq \frac{-1}{8}\Leftrightarrow x\neq \frac{-1}{2}\)
Vậy TXĐ \(D=\mathbb{R}\setminus \left\{\frac{-1}{2}\right\}\)
d)
ĐK:
\(|x-2015|+1\neq 0\Leftrightarrow |x-2015|\neq -1\Leftrightarrow x\in\mathbb{R}\)
Vậy TXĐ \(D=\mathbb{R}\)
e)
ĐK: \(\left\{\begin{matrix} |x-1,2|\neq 0\\ 2x-5\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\neq 1,2\\ x\neq 2,5\end{matrix}\right.\)
Vậy TXĐ: \(D=\mathbb{R}\setminus \left\{1,2; 2,5\right\}\)
f)
ĐK: \(x^2-4\neq 0\Leftrightarrow (x-2)(x+2)\neq 0\Leftrightarrow x\neq \pm 2\)
Vậy TXĐ: \(D=\mathbb{R}\setminus \left\{\pm 2\right\}\)
Tìm x, y, z biết:
a) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và 2x + 3y + z = 17
Giải
Ta có: \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{z}{4}\) và 2x + 3y + z = 17
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{z}{4}=\dfrac{2x+3y+z}{4+9+4}=\dfrac{17}{17}=1\)
\(\dfrac{x}{2}=1\Rightarrow x=2\)
\(\dfrac{y}{3}=1\Rightarrow y=3\)
\(\dfrac{z}{4}=1\Rightarrow z=4\)
Vậy...
b) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và (x - y)2 + (y - z)2 = 2
Giải
Áp dụng tính chất của dãy tỉ số bằng nhau:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{\left(x-y\right)^2+\left(y-z\right)^2}{\left(2-3\right)^2+\left(3-4\right)^2}=\dfrac{2}{2}=1\)
\(\dfrac{x}{2}=1\Rightarrow x=2\)
\(\dfrac{y}{3}=1\Rightarrow y=3\)
\(\dfrac{z}{4}=1\Rightarrow z=4\)
Vậy...
Câu 1:
a: |x-1|+|x-5|=0
\(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
b: Đặt \(\dfrac{x}{3}=\dfrac{y}{2}=k\)
=>x=3k; y=2k
\(x^2+y^2=52\)
\(\Rightarrow9k^2+4k^2=52\)
\(\Leftrightarrow k^2=4\)
Trường hợp 1: k=2
=>x=6; y=4
Trường hợp 2: k=-2
=>x=-6; y=-4
c: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a}{2a-b}=\dfrac{bk}{2bk-b}=\dfrac{k}{2k-1}\)
\(\dfrac{c}{2c-d}=\dfrac{dk}{2dk-d}=\dfrac{k}{2k-1}\)
Do đó: \(\dfrac{a}{2a-b}=\dfrac{c}{2c-d}\)
1)
x(x-y) = \(\dfrac{3}{10}\)
=> \(x^2-xy=\dfrac{3}{10}\) (1)
y(x-y) = \(-\dfrac{3}{50}\)
=> \(xy-y^2=-\dfrac{3}{50}\) (2)
Trừ (1) cho (2), ta có :
\(x^2-xy-xy+y^2=\dfrac{3}{10}+\dfrac{3}{50}\)
\(\Rightarrow x^2-2xy+y^2=\dfrac{18}{50}=\dfrac{9}{25}\)
=> \(\left(x-y\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}x-y=\dfrac{3}{5}\\x-y=-\dfrac{3}{5}\end{matrix}\right.\)
TH1
x- y = \(\dfrac{3}{5}\)
Ta có
\(\left\{{}\begin{matrix}x\left(x-y\right)=\dfrac{3}{10}\\y\left(x-y\right)=-\dfrac{3}{50}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{5}x=\dfrac{3}{10}\\\dfrac{3}{5}y=-\dfrac{3}{50}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=-\dfrac{1}{10}\end{matrix}\right.\)
TH2:
x-y=\(-\dfrac{3}{5}\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(x-y\right)=\dfrac{3}{10}\\y\left(x-y\right)=-\dfrac{3}{50}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-\dfrac{3}{5}x=\dfrac{3}{10}\\-\dfrac{3}{5}y=-\dfrac{3}{50}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{5}\end{matrix}\right.\)
Vậy các cặp (x,y) thỏa mãn là (x;y) \(\in\left\{\left(\dfrac{1}{2};-\dfrac{1}{5}\right);\left(-\dfrac{1}{2};\dfrac{1}{5}\right)\right\}\)
2) \(\left(x-3\right)\left(x+\dfrac{1}{2}\right)>0\)
TH1:
\(\left\{{}\begin{matrix}x-3>0\\x+\dfrac{1}{2}>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>3\\x>-\dfrac{1}{2}\end{matrix}\right.\)
=> x >3
TH2:
\(\left\{{}\begin{matrix}x-3< 0\\x+\dfrac{1}{2}< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< 3\\x< -\dfrac{1}{2}\end{matrix}\right.\)
=> x <\(-\dfrac{1}{2}\)
Vậy giá trị x thỏa mãn là x < -1/2 hoặc x>3
1)
Từ gt,ta có : x(x - y) - y(x - y) =\(\frac{3}{10}-\frac{-3}{50}\)
(x - y)2 =\(\frac{9}{25}\)\(\Rightarrow\orbr{\begin{cases}x-y=\frac{3}{5}\\x-y=\frac{-3}{5}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}:\frac{3}{5}=\frac{1}{2}\\x=\frac{3}{10}:\frac{-3}{5}=\frac{-1}{2}\end{cases};\orbr{\begin{cases}y=\frac{-3}{50}:\frac{3}{5}=\frac{-1}{10}\\y=\frac{-3}{50}:\frac{-3}{5}=\frac{1}{10}\end{cases}}}}\)
Vậy\(x=\frac{1}{2};y=\frac{-1}{10}\) hoặc\(x=\frac{-1}{2};y=\frac{1}{10}\)
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
\(\Rightarrow x;1-2y\in U\left(40\right)\)
\(U\left(40\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)
Mà 1-2y lẻ nên:
\(\left\{{}\begin{matrix}1-2y=1\Rightarrow2y=0\Rightarrow y=0\\x=40\\1-2y=-1\Rightarrow2y=2\Rightarrow y=1\\x=-40\end{matrix}\right.\)
\(\left\{{}\begin{matrix}1-2y=5\Rightarrow2y=-4\Rightarrow y=-2\\x=8\\1-2y=-5\Rightarrow2y=6\Rightarrow y=3\\x=-8\end{matrix}\right.\)
b tương tự.
c) \(\left(x+1\right)\left(x-2\right)< 0\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-2>0\Rightarrow x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-2< 0\Rightarrow x< 2\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow-1< x< 2\Rightarrow x\in\left\{0;1\right\}\)
d tương tự