DT
2
K
Khách
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DT
3
XO
26 tháng 6 2023
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne4\end{matrix}\right.\)
\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)
\(\Leftrightarrow\left(x-3\right).\left(x-4\right)+\left(x-2\right)^2=-\left(x-2\right).\left(x-4\right)\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow\left(3x-8\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=3\end{matrix}\right.\left(\text{thỏa}\right)\)
DT
Dang Tung
CTVHS
26 tháng 6 2023
\(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\left(x\ne\left\{2;4\right\}\right)\\ =>\dfrac{\left(x-3\right)\left(x-4\right)+\left(x-2\right)^2}{\left(x-2\right)\left(x-4\right)}=-1\\ =>x^2-3x-4x+12+x^2-4x+4=-\left(x-2\right)\left(x-4\right)\\ =>2x^2-11x+16=-x^2+6x-8\\ =>3x^2-17x+24=0\\ =>\left(x-3\right)\left(3x-8\right)=0\\ =>\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\left(TMDK\right)\)
TH
4
TH
2
18 tháng 9 2023
TA CÓ 0=02
⇒X-11+Y+X+4-Y=0
⇒(X+X)+(-11+4)+(Y-Y)=0
⇒2X+(-7)+0=0
⇒2X=0-(-7)
⇒2X=7
⇒X=7:2
⇒X=3,5
VẬY X =3,5
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 8 2023
\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)
\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)
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}
⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)
NX
9 tháng 8 2023
\(\dfrac{x}{9}< \dfrac{4}{7}< \dfrac{x+1}{9}\)
=>\(\dfrac{7x}{63}< \dfrac{36}{63}< \dfrac{7x+7}{63}\)
\(\Rightarrow7x< 36< 7x+7\)
\(\Rightarrow x< \dfrac{36}{7}< x+1\)
\(\Rightarrow x< 5\dfrac{1}{7}< x+1\)
\(\Rightarrow x=5\)
13 tháng 3 2023
x/y=3/4
=>x/3=y/4
=>x/15=y/20
y/z=5/7
=>y/5=z/7
=>y/20=z/28
=>x/15=y/20=z/28=(2x+3y-z)/(2*15+3*20-28)=186/62=3
=>x=45; y=60; z=84
T
9 tháng 8 2023
a) \(\dfrac{49}{81}=\dfrac{7^x}{9^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{7}{9}\right)^2=\left(\dfrac{7}{9}\right)^x\)\(\Rightarrow x=2\)
b) \(\dfrac{-64}{343}=\left(-\dfrac{4^x}{7^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{4}{7}\right)^3=\left(-\dfrac{4}{7}\right)^x\) \(\Rightarrow x=3\)
c) \(\dfrac{9}{144}=\dfrac{3^x}{12^x}\)(sửa đề)
\(\Leftrightarrow\left(\dfrac{3}{12}\right)^2=\left(\dfrac{3}{12}\right)^x\Rightarrow x=2\)
d) \(-\dfrac{1}{32}=\left(-\dfrac{1^x}{2^x}\right)\)(sửa đề)
\(\Leftrightarrow\left(-\dfrac{1}{2}\right)^5=\left(-\dfrac{1}{2}\right)^x\Rightarrow x=5\)
Mong bạn xem lại đề bài.
DT
13
VN
Võ Ngọc Phương
CTVHS
VIP
26 tháng 8 2023
a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)
\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)
\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)
\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)
\(x=\dfrac{-1}{3}\)
b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)
\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)
\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)
\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)
\(2x=\dfrac{-17}{24}\)
\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)
\(x=\dfrac{-17}{48}\)
c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)
26 tháng 8 2023
a, 1/4 + 3/4 : x = -2
3/4 : x = -2 - 1/4
3/4 : x = -9/4
x = 3/4 : -9/4
x = -1/3
16 tháng 11 2021
Bài 2:
\(\dfrac{1}{2}:\dfrac{5}{4}=x:\dfrac{10}{3}\Leftrightarrow\dfrac{1}{2}.\dfrac{4}{5}=\dfrac{3}{10}x\Leftrightarrow\dfrac{3}{10}x=\dfrac{2}{5}\Leftrightarrow x=\dfrac{2}{5}:\dfrac{3}{10}=\dfrac{4}{3}\)
Bài 3:
Áp dụng t/c dtsbn:
\(\dfrac{x}{4}=\dfrac{y}{12}=\dfrac{x+y}{4+12}=\dfrac{48}{16}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.4=12\\y=3.12=36\end{matrix}\right.\)
\(x^2\) + (\(x\) + 2)(11\(x\) - 7) = 4
(\(x\) + 2)(11\(x\) - 7) + \(x^2\) - 4 = 0
(\(x+2\))(11\(x-7\)) + (\(x-2\))(\(x+2\)) = 0
(\(x+2\))(11\(x\) - 7 + \(x-2\)) = 0
(\(x\) + 2)(12\(x\) - 9) = 0
\(\left[{}\begin{matrix}x+2=0\\12x-9=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2\\x=\dfrac{9}{12}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
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