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a) 7.28=x.x
=> 196=x2
=> \(\left(\pm14\right)^2=x^2\)
=> x=\(\pm14\)
b) DK: x≠-17
pt<=> 4.(10+2)=6.(17+x)
=> 4.12=17.6+6x
=> 48-102=6x
=>-66=6x
=>x=-11
c) 7.(x+40)=6.(17+x)
=> 7x+280=102+6x
=> 7x-6x=102-280
=> x=-178
Giải:
a) \(\dfrac{7}{x}=\dfrac{x}{28}\)
\(\Leftrightarrow x^2=196\)
\(\Leftrightarrow x=\pm\sqrt{196}=\pm14\)
Vậy ...
b) \(\dfrac{10+2}{17+x}=\dfrac{3}{4}\)
\(\Leftrightarrow40+8=51+3x\)
\(\Leftrightarrow3x=40+8-51=-3\)
\(\Leftrightarrow x=-\dfrac{3}{3}=-1\)
Vậy ...
c) \(\dfrac{40+x}{17+x}=\dfrac{6}{7}\)
\(\Leftrightarrow280+7x=102+6x\)
\(\Leftrightarrow7x-6x=102-280\)
\(\Leftrightarrow x=-178\)
Vậy ...
1.a) Dễ nhận thấy đề toán chỉ giải được khi đề là tìm x,y. Còn nếu là tìm x ta nhận thấy ngay vô nghiệm. Do đó: Sửa đề: \(\left|x-3\right|+\left|2-y\right|=0\)
\(\Leftrightarrow\left|x-3\right|=\left|2-y\right|=0\)
\(\left|x-3\right|=0\Rightarrow\left\{{}\begin{matrix}x-3=0\\-\left(x-3\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\) (1)
\(\left|2-y\right|=0\Rightarrow\left\{{}\begin{matrix}2-y=0\\-\left(2-y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\) (2)
Từ (1) và (2) có: \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=3\\x_2=-3\end{matrix}\right.\\\left\{{}\begin{matrix}y_1=2\\y_2=-2\end{matrix}\right.\end{matrix}\right.\)
a: (x+1/2)(2/3-2x)=0
=>x+1/2=0 hoặc 2/3-2x=0
=>x=-1/2 hoặc x=1/3
b: 
c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)
\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)
\(\dfrac{-8}{13}+\dfrac{-7}{17}+\dfrac{21}{13}\le x\le\dfrac{-9}{14}+3+\dfrac{5}{-14}\)
=> \(\dfrac{10}{17}\le x\le2\)
=> \(\dfrac{10}{17}\le\dfrac{17x}{17}\le\dfrac{34}{17}\)
=> 10 \(\le17x\le34\)
=> x = 1; 2 (thỏa mãn)
@Khánh Linh
7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)
\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)
\(\Rightarrow-2\le x\le2\)
\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\)
\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)
\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)
\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)
\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)
\(\Rightarrow x\in\left\{11;12;13;14\right\}\)
8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)
tìm x:
\(5\dfrac{8}{17}:x+\left(-\dfrac{4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)
Cho link đây, hồi chiều mới làm: Câu hỏi của nguyen thu trang - Toán lớp 6 | Học trực tuyến
(5\(\dfrac{8}{17}\)+\(\dfrac{-4}{17}\)): x +\(\dfrac{22}{7}\):\(\dfrac{52}{3}\)=\(\dfrac{4}{11}\)
\(^{\dfrac{89}{17}}\):x+\(\dfrac{22}{7}\)x\(\dfrac{3}{52}\)=\(\dfrac{4}{11}\)
\(\dfrac{89}{17}\):x+\(\dfrac{33}{182}\)=\(\dfrac{4}{11}\)
\(\dfrac{89}{17}\):x=\(\dfrac{385}{2002}\)
x=\(\dfrac{2314}{85}\)
\(\dfrac{9}{17}\times\dfrac{21}{13}+\dfrac{9}{17}\times\dfrac{5}{13}-\dfrac{9}{17}\times2\)
\(=\dfrac{9}{17}\times\left(\dfrac{21}{13}+\dfrac{5}{13}-2\right)\)
\(=\dfrac{9}{17}\times\left(\dfrac{26}{13}-2\right)=\dfrac{9}{17}\times\left(2-2\right)\)
\(=\dfrac{9}{17}\times0=0\)
\(\dfrac{x-17}{1990}+\dfrac{x-21}{1986}+\dfrac{x+1}{1004}=4\)
\(\Rightarrow\left(\dfrac{x-17}{1990}-1\right)+\left(\dfrac{x-21}{1986}-1\right)+\left(\dfrac{x+1}{1004}-2\right)=0\)
\(\Rightarrow\dfrac{x-2007}{1990}+\dfrac{x-2007}{1986}+\dfrac{x-2007}{1004}=0\)
\(\Rightarrow\left(x-2007\right)\left(\dfrac{1}{1990}+\dfrac{1}{1986}+\dfrac{1}{1004}\right)=0\)
\(\Rightarrow x-2007=0\Rightarrow x=2007\)
Vậy x = 2007
tks nhìu nhan