(1 điểm) Tìm $x$, biết: $|x-2,4|=\dfrac{1}{2}$.
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\(\dfrac{2}{3}-\left|x-2,4\right|=\dfrac{1}{2}\)
\(\left|x-2,4\right|=\dfrac{2}{3}-\dfrac{1}{2}\)
\(\left|x-2,4\right|=\dfrac{1}{6}\)
*) Với \(x\ge2,4\) ta có:
\(x-2,4=\dfrac{1}{6}\)
\(x=\dfrac{1}{6}+2,4\)
\(x=\dfrac{77}{30}\) (nhận)
*) Với \(x< 2,4\) ta có:
\(x-2,4=-\dfrac{1}{6}\)
\(x=-\dfrac{1}{6}+2,4\)
\(x=\dfrac{67}{30}\) (nhận)
Vậy \(x=\dfrac{67}{30};x=\dfrac{77}{30}\)
a, \(\dfrac{3}{7}\)\(x\) - 0,4 = - \(\dfrac{17}{35}\)
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{17}{35}\) + 0,4
\(\dfrac{3}{7}\)\(x\) = - \(\dfrac{3}{35}\)
\(x\) = - \(\dfrac{3}{35}\): \(\dfrac{3}{7}\)
\(x\) = - \(\dfrac{1}{5}\)
b, 0,2.(\(x\) - 3) +2,4 = 10
0,2.(\(x\) - 3) = 10 - 2,4
0,2.(\(x\) - 3) = 7,6
\(x\) - 3 = 7,6:0,2
\(x\) - 3 = 38
\(x\) = 38 + 3
\(x\) = 41
\(a,5,2x+7\dfrac{2}{5}=6\dfrac{3}{4}\\ \Rightarrow\dfrac{26}{5}x+\dfrac{37}{5}=\dfrac{27}{4}\\ \Rightarrow\dfrac{26}{5}x=-\dfrac{13}{20}\\ \Rightarrow x=-\dfrac{1}{8}\\ b,2,4:\left(\dfrac{-1}{2}-x\right)=1\dfrac{3}{5}\\ \Rightarrow\dfrac{12}{5}:\left(\dfrac{-1}{2}-x\right)=\dfrac{8}{5}\\ \Rightarrow\dfrac{-1}{2}-x=\dfrac{3}{2}\\ \Rightarrow x=-2\)
a: Sửa đề: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
Để A là số nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3+4⋮\sqrt{x}-3\)
=>\(4⋮\sqrt{x}-3\)
=>\(\sqrt{x}-3\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)
=>\(\sqrt{x}\in\left\{4;2;5;1;7\right\}\)
=>\(x\in\left\{16;4;25;1;49\right\}\)
b:
Bài 1: Ta có: \(4\dfrac{3}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{23}{5}+\dfrac{7}{10}< X< \dfrac{20}{3}\)
\(\dfrac{138}{30}< X< \dfrac{200}{3}\)
\(\Rightarrow X\in\left\{\dfrac{160}{30};\dfrac{161}{30};\dfrac{162}{30};...;\dfrac{198}{30};\dfrac{199}{30}\right\}\)
Bài 2: \(X-2019\dfrac{2}{13}=3\dfrac{7}{26}+4\dfrac{7}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{85}{26}+\dfrac{215}{52}\)
\(\Rightarrow X-\dfrac{26249}{13}=\dfrac{385}{52}\)
\(\Rightarrow X=\dfrac{105381}{52}\)
Các đa thức là: \( - {x^2} + 3x + 1;\dfrac{x}{{\sqrt 5 }};2024;3{x^2}{y^2} - 5{x^3}y + 2,4.\)
\(1.x-\dfrac{2}{3}\times\left(x+9\right)=1\)
\(x-\dfrac{2}{3}\times x-6=1\)
\(x\times\left(1-\dfrac{2}{3}\right)=7\)
\(x\times\dfrac{1}{3}=7\)
\(x=21\)
\(2.x-\dfrac{11}{15}=\dfrac{3+x}{5}\)
\(\dfrac{15x}{15}-\dfrac{11}{15}=\dfrac{9+3x}{15}\)
\(15x-11=9+3x\)
\(12x=20\)
\(x=\dfrac{5}{3}\)
a) \(\left|x-1\right|=5\)
\(\Rightarrow\orbr{\begin{cases}x-1=5\\x-1=-5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=6\\x=-4\end{cases}}\)
b) \(\left|x-0,5\right|=2,4\)
\(\Rightarrow\orbr{\begin{cases}x-0,5=2,4\\x-0,5=-2,4\end{cases}}\Rightarrow\orbr{\begin{cases}x=2,9\\x=-1,9\end{cases}}\)
c) \(2\left|3x-1\right|-1=6\)
\(\Leftrightarrow2\left|3x-1\right|=7\)
\(\Leftrightarrow\left|3x-1\right|=3,5\)
\(\Rightarrow\orbr{\begin{cases}3x-1=3,5\\3x-1=-3,5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}3x=4,5\\2x=-2,5\end{cases}\Rightarrow\orbr{\begin{cases}x=1,5\\-1,25\end{cases}}}\)
a: ĐKXĐ: x<>0; x<>1
\(P=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |2x+1|=3
=>x=1(loại); x=-2(nhận)
Khi x=-2 thì P=4/-3=-4/3
c: P=-1/2
=>x^2/x-1=-1/2
=>2x^2=-x+1
=>2x^2+x-1=0
=>2x^2+2x-x-1=0
=>(x+1)(2x-1)=0
=>x=1/2; x=-1
| x - 2,4| = \(\dfrac{1}{2}\)
\(\left[{}\begin{matrix}x-2,4=\dfrac{1}{2}(đk:x>2,4)\\x-2,4=-\dfrac{1}{2}(đk:x< 2,4)\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{1}{2}+2,4\\x=-\dfrac{1}{2}+2,4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2,9(tm)\\x=1,9(tm)\end{matrix}\right.\)
vậy \(x\in\) { 1,9 ; 2,9}
x=2/9