giúp mình phần tìm x và tìm a,b với ạ
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a) 2x - 3 = x + 1/2
<=> 2x - 3 = 1/2x + 1/2
<=> 2x - 3 - 1/2x = 1/2
<=> 3/2x - 3 = 1/2
<=> 3/2x = 1/2 + 3
<=> 3/2x = 7/2
<=> x = 7/2 : 3/2
<=> x = 7/3
=> x = 7/3
\(a,2x-3=x+\frac{1}{2}\)
\(2x-3=\frac{1}{2}x+\frac{1}{2}\)
\(2x-3-\frac{1}{2}x=\frac{1}{2}\)
\(\frac{3}{2}x-3=\frac{1}{2}\)
\(\frac{3}{2}x=\frac{1}{2}+3\)
\(\frac{3}{2}x=\frac{7}{2}\)
\(x=\frac{7}{2}:\frac{3}{2}\)
\(x=\frac{14}{6}=\frac{7}{3}\)
\(\)B làm tương tự
a.
\(A=B\)
\(\Leftrightarrow\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{-16}{x^2-4}\);ĐK:\(x\ne\pm2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{-16}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=-16\)
\(\Leftrightarrow x^2+4x+4-x^2+4x-4+16=0\)
\(\Leftrightarrow8x+16=0\)
\(\Leftrightarrow8\left(x+2\right)=0\)
\(\Leftrightarrow x=-2\left(ktm\right)\)
Vậy không có giá trị x thỏa mãn A=B
b.
\(A:B=\dfrac{\left(x+2\right)^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{-16}{\left(x-2\right)\left(x+2\right)}< 0\)
\(\Leftrightarrow\dfrac{x^2+4x+4-x^2+4x-4}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{-16}< 0\)
\(\Leftrightarrow\dfrac{8x}{16}>0\)
\(\Leftrightarrow\dfrac{x}{2}>0\)
\(\Leftrightarrow x>0\)
a) 16 = 2⁴
42 = 2.3.7
ƯCLN(16; 42) = 2
ƯC(16; 42) = Ư(2) = {1; 2}
b) 16 = 2⁴
42 = 2.3.7
86 = 2.43
ƯCLN(16; 42; 86) = 2
ƯC(16; 42; 86) = Ư(2) = {1; 2}
c) 25 = 5²
75 = 3.5²
ƯCLN(25; 75) = 5² = 25
ƯC(25; 75) = Ư(25) = {1; 5; 25}
d) 25 = 5²
55 = 5.11
75 = 3.5²
ƯCLN(25; 55; 75) = 5
ƯC(25; 55; 75) = Ư(5) = {1; 5}
\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
7\(x\) < 36 < 63\(x\) + 7
⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)
\(\dfrac{29}{63}\)< \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)
=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)
\(\Rightarrow7x< 36< 7x+7\)
\(\Rightarrow x< \dfrac{36}{7}< x+1\)
\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)
\(\Rightarrow x=5\)
Câu 3:
\(a,\dfrac{a}{2}=\dfrac{b}{5}=\dfrac{a-b}{2-5}=\dfrac{9}{-3}=-3\\ \Rightarrow\left\{{}\begin{matrix}a=-6\\b=-15\end{matrix}\right.\\ b,\dfrac{a}{3}=\dfrac{b}{6}=\dfrac{a-b}{3-6}=\dfrac{12}{-3}=-4\\ \Rightarrow\left\{{}\begin{matrix}a=-12\\b=-24\end{matrix}\right.\\ c,\dfrac{a}{4}=\dfrac{b}{5}=\dfrac{a-2b}{4-10}=\dfrac{-30}{-6}=5\\ \Rightarrow\left\{{}\begin{matrix}a=20\\b=25\end{matrix}\right.\)