giúp mik bài này với
a) |x - 2| + |y + 3| = 0
b) |x - 2| - |x + 3| = 0
c) |x - 3/4| + |x + 5/4| = 1Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dễ mak
nhưng mik nhìn đề thấy dài quá nên ko muốn làm
hihi^_$
Bài 1 :
a, \(\left(x-3\right)^2-4=0\Leftrightarrow\left(x-3\right)^2=4\Leftrightarrow\left(x-3\right)^2=\left(\pm2\right)^2\)
TH1 : \(x-3=2\Leftrightarrow x=5\)
TH2 : \(x-3=-2\Leftrightarrow x=1\)
b, \(x^2-2x=24\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)
TH1 : \(x-6=0\Leftrightarrow x=6\)
TH2 : \(x+4=0\Leftrightarrow x=-4\)
c, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-4\right)=0\)
\(\Leftrightarrow2x+30=0\Leftrightarrow x=-15\)
d, tương tự
a. \(\left(x^2-2x+1\right)-3x\left(x-1\right)=0\)
\(\Leftrightarrow x^2-2x+1-3x^2+3x=0\)
\(\Leftrightarrow-2x^2+x+1=0\)
\(\Leftrightarrow-2x^2+2x-x+1=0\)
\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow-\left(2x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)
Vậy \(x\in\left\{-\frac{1}{2};1\right\}\)
b. \(4\left(7x-3\right)-\left(7x^2-3x\right)=0\)
\(\Leftrightarrow4\left(7x-3\right)-x\left(7x-3\right)=0\)
\(\Leftrightarrow\left(4-x\right)\left(7x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\7x-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{3}{7}\end{cases}}\)
Vậy \(x\in\left\{4;\frac{3}{7}\right\}\)
c.\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)
\(\Leftrightarrow\left(5-x\right)\left(2+3x\right)=\left(2-3x\right)\left(2+3x\right)\)
\(\Leftrightarrow\left(2+3x\right)\left(5-x-2+3x\right)=0\)
\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=0\\2x+3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)
Vậy \(x\in\left\{-\frac{2}{3};-\frac{3}{2}\right\}\)
d. \(7-\left(2x+4\right)=-\left(x+4\right)\)
\(\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow7-4+4=-x+2x\)
\(\Leftrightarrow7=x\)
Vậy x = 7
e. \(\left(x-1\right)-\left(2x-1\right)=9\)
\(\Leftrightarrow x-1-2x+1=9\)
\(\Leftrightarrow-x=9\)
\(\Leftrightarrow x=-9\)
g. \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}}\)Mà : \(x^2+1\ge1>0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy x = -1
Bài 2 :
a, \(\left|x-\frac{5}{3}\right|< \frac{1}{3}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{3}< \frac{1}{3}\\x-\frac{5}{3}< -\frac{1}{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 2\\x< \frac{4}{3}\end{cases}}}\)
b, \(\frac{2}{5}< \left|x-\frac{7}{5}\right|< \frac{3}{5}\)
\(\orbr{\begin{cases}\frac{2}{5}< x-\frac{7}{5}< \frac{3}{5}\\\frac{2}{5}< -x+\frac{7}{5}< \frac{3}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{9}{5}< x< 2\\1>x>\frac{4}{5}\end{cases}}\)
a) Ta có :
\(\left|\frac{3}{4}x-4\right|\ge0\)
\(\left|3x+5\right|\ge0\)
\(\Rightarrow\left|\frac{3}{4}x-4\right|+\left|3x+5\right|\ge0\)
Mà : \(\left|\frac{3}{4}x-4\right|+\left|3x+5\right|=0\) (đề bài)
\(\Rightarrow\hept{\begin{cases}\frac{3}{4}x-4=0\\3x+5=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{16}{3}\\x=-\frac{5}{3}\end{cases}}\)
Vì trong một phương trình không thể cùng có 2 giá trị
=> Không có giá trị x thõa mãn đề bài
\(\left(x-3\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=12\end{cases}}\)
\(\Rightarrow x\in\left\{3;12\right\}\)
\(\left(x^2-81\right)\left(x^2+9\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-81=0\\x^2+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x\in\varnothing\end{cases}}\Leftrightarrow x=9\)
\(\Rightarrow x=9\)
\(\left(x-4\right)\left(x+2\right)< 0\)
\(\Rightarrow\hept{\begin{cases}x-4\\x+2\end{cases}}\)trái dấu
\(TH1:\hept{\begin{cases}x-4>0\\x+2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x< -2\end{cases}}\Leftrightarrow x\in\varnothing\)
\(TH2:\hept{\begin{cases}x-4< 0\\x+2>0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x< 4\\x>-2\end{cases}}\Leftrightarrow x\in\left\{-1;0;1;2;3\right\}\)
Vậy \(x\in\left\{-1;0;1;2;3\right\}\)
1a) x.y = -15 = (-3).5 = (-5).3 = (-1).15 = (-15).1
Vậy x = { -3;5;-5;3;-1;15;-15;1}
Với y tương ứng = { 5;-3;3;-5;15;-1;1;-15}
b) x.y = -13 = (-1).13 = (-13).1
Vậy x = { -1;13;-13;1}
Với y tương ứng = { 13;-1;1;-13}
c) x.y = 85 = 1.85 = 85.1 = 5.17 = 17.5
Vậy x = {1;85;85;1;5;17;17;5}
Với y tương ứng = { 85;1;1;85;17;5;5;17}
2;3: Tự làm
a) \(...\Rightarrow\left\{{}\begin{matrix}x-2=0\\y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)
b) \(...\Rightarrow|x-2|=|x+3|\Rightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-x-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0x=5\\2x=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x=-\dfrac{1}{2}\)
c) \(|x-\dfrac{3}{4}|+|x+\dfrac{5}{4}|=1\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{3}{4}\le0\\x+\dfrac{5}{4}\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{4}\\x\ge-\dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow-\dfrac{5}{4}\le x\le\dfrac{3}{4}\)