Tìm x thuộc Z biết
l x+19 l + l x+5 l + l x+20.11 l =4x
Làm giúp mình với ngày mai có kiểm tra toán rồi
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Ta co : \(\left|x+25\right|\ge0\forall x\in Z\)
\(\left|-y+5\right|\ge0\forall x\in Z\)
Mà : |x + 25| + |-y + 5| = 0
Nên : \(\hept{\begin{cases}\left|x+25\right|=0\\\left|-y+5\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x+25=0\\-y+5=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=-25\\y=5\end{cases}}\)
a. |x + 8| = 6
TH1: x + 8 = -6
x = -14
TH2: x + 8 = 6
x = -2
b. 1 < |x - 2| < 4
\(\Rightarrow\left|x-2\right|\in\left\{2;3\right\}\)
TH1: |x - 2| = 2
\(\Leftrightarrow\orbr{\begin{cases}x-2=-2\\x-2=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
TH2: |x - 2| = 3
\(\Leftrightarrow\orbr{\begin{cases}x-2=-3\\x-2=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
c. |x - a| = a
TH1: x - a = -a
x = 0
TH2: x - a = a
x = 2a
a, =) x+6= -6 hc 6 =) +) x+6=6=)x=0 +) x+6= -6 =)x=-1 mk lm dc mỗi í a thôi à :) chúc pn hok tốt
a)|2x-5|=13
2x-5=13=>x=9
2x-5=-13=>x=-4
b)3|x+1|+1=28
3|x+1|=28-1
3|x+1|=27
|x+1|=27:3
|x+1|=9
x+1=9=>x=8
x+1=-9=>x=-10
tick nha
a)(x+1)+(x+3)+...+(x+97)+(x+99)=0
x.50+2500=0
x.50=0-2500
x.50=-2500
x=-2500:5
x=-500
\(1)|5-2x|=|x+4|\)
\(\Leftrightarrow\orbr{\begin{cases}5-2x=x+4\\5-2x=-x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x-x=4-5\\-2x+x=-4-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}-3x=-1\\-x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=9\end{cases}}}\)
Vậy \(x=\frac{1}{3};x=9\)
\(2)|x-1|=|2x+5|\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=2x+5\\x-1=-2x-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2x=5+1\\x+2x=-5+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}-x=4\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{4}{3}\end{cases}}}\)
Vậy \(x=-4;x=-\frac{4}{3}\)
\(3)|x+1|+|x+2|+|x+3|=0\left(1\right)\)
Ta có: \(|x+1|\ge0\forall x;|x+2|\ge0\forall x;|x+3|\ge0\forall x\)
\(\Leftrightarrow|x+1|+|x+2|+|x+3|\ge0\forall x\)
\(\left(1\right)\Leftrightarrow|x+1|+|x+2|+|x+3|=0\)
\(\Leftrightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=0\)
\(\Leftrightarrow x+1+x+2+x+3=0\)
\(\Leftrightarrow\left(x+x+x\right)+\left(1+2+3\right)=0\)
\(\Leftrightarrow3x+6=0\)
\(\Leftrightarrow3x=-6\)
\(\Leftrightarrow x=-6:3\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
a) Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0\forall x\in Q\)
\(\left|y+\dfrac{2017}{2018}\right|\ge0\forall y\in Q\)
\(\left|z-2019\right|\ge0\forall x\in Q\)
\(\Rightarrow\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|\ge0\forall x,y,z\in Q\)
Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\).
b) Lại có:
\(\left|x-\dfrac{9}{5}\right|\ge0\forall x\in Q\)
\(\left|y+\dfrac{3}{4}\right|\ge0\forall y\in Q\)
\(\left|z+\dfrac{7}{2}\right|\ge0\forall z\in Q\)
\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,zQ\)
Mà theo đề bài:
\(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\forall\)
\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{9}{5}\right|=0\\\left|y+\dfrac{3}{4}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy .....
a) \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\)
Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0;\left|y+\dfrac{2017}{2018}\right|\ge0;\left|z-2019\right|\ge0\)
Để \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\) thì:
\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\)
Vậy............................
b) Ta có: \(\left|x-\dfrac{9}{5}\right|\ge0;\left|y+\dfrac{3}{4}\right|\ge0;\left|z+\dfrac{7}{2}\right|\ge0\)
Mà \(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\) thì:
\(\left|x-\dfrac{9}{5}\right|=\left|y+\dfrac{3}{4}\right|=\left|z+\dfrac{7}{2}\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)
Vậy............................
a) ĐKXĐ : \(x\ne0;x\ne5\)
b) \(A=\frac{x}{x-5}-\frac{10}{x}-\frac{75}{5x-x^2}\)
\(A=\frac{x}{x-5}-\frac{10}{x}+\frac{75}{x^2-5x}\)
\(A=\frac{x^2-10\left(x-5\right)+75}{x\left(x-5\right)}=\frac{x^2-10x+25}{x\left(x-5\right)}=\frac{\left(x-5\right)^2}{x\left(x-5\right)}=\frac{x-5}{x}\)
c) \(A=2\Leftrightarrow\frac{x-5}{x}=2\Leftrightarrow x-5=2x\Leftrightarrow x=-5\) (t/m)
d) \(A=\frac{x-5}{x}=1-\frac{5}{x}\)
\(A\in Z\Leftrightarrow\) \(x\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\). Kết hợp đkxđ => \(x\in\left\{-5;-1;1\right\}\) tmđb
\(\text{ -2005 < l x +5 l _< 1}\text{ -2005 < l x +5 l _< 1}\) -2005 < l x +5 l \(\le\)1
xét l x +5 l \(\ge\)0
mà theo đề bài thì l x +5 l \(\le\)1
nên l x +5 l = 1 hoặc 0
nếu l x +5 l = 1
=) x +5 = 1
=) x = 1 - 5 = -4
nếu l x +5 l = 0
=) x +5 = 0
=) x = 0 - 5 = -5
=) \(x\in\left\{-5;-4\right\}\)
đăng kí kênh của V-I-S nha !
ĐK : 4x \(\ge\)0 => x\(\ge\)0
=> |x + 19| \(\ge\)0 ; |x + 5| \(\ge\)0 ; |x + 20.11| \(\ge\)0
Khi đó :
x + 19 + x + 5 + x + 20.11 = 4x
=> 3x + 19 + 5 + 220 = 4x
=> 3x + 244 = 4x
=> 244 = 4x - 3x
=> 224 = x
=> x = 224 (t/m Đk)
cmn lỗi đánh máy : x = 244 nha >: