Question

tìm x biết

a) \(\dfrac{x}{27}=\dfrac{-3}{x}\)

b) \(\dfrac{-9}{x}=\dfrac{-x}{\dfrac{4}{49}}\)

c)\(\left|7x-\dfrac{5}{3}\right|+\dfrac{7}{19}=\dfrac{-8}{15}\)
d)\(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=\dfrac{18}{19}-1\dfrac{2}{5}\)

Answer 1

Mình chỉ giải câu a thôi,mấy câu còn lại dễ.

a)Ta có:\(\dfrac{x}{27}=\dfrac{-3}{x}\)

=>\(x^2=-3\cdot27=-81\)(Nhân chéo)

Mà x²>0 với mọi x nên :

Không có giá trị nào thỏa mãn điều kiện của x

Answer 2

Tìm x biết :

a) \(\dfrac{x}{27}=-\dfrac{3}{x}\) \(\Rightarrow2x=-3.27\Rightarrow2x=-81\Rightarrow x=-40,5\)

b) \(-\dfrac{9}{x}=-\dfrac{x}{\dfrac{4}{49}}\Rightarrow2x=-9.\left(-\dfrac{4}{9}\right)\Rightarrow2x=4\Rightarrow x=2\)

c) \(\left|7x-\dfrac{5}{3}\right|+\dfrac{7}{19}=-\dfrac{8}{15}\) ( mk nghĩ bn chép sai đề bài câu này )

\(\Rightarrow\left|7x-\dfrac{5}{3}\right|=-\dfrac{8}{15}-\dfrac{7}{19}\)

\(\Rightarrow\left|7x-\dfrac{5}{3}\right|=-\dfrac{257}{285}\)

\(\Rightarrow\left[{}\begin{matrix}7x-\dfrac{5}{3}=-\dfrac{257}{285}\\7x-\dfrac{5}{3}=\dfrac{257}{285}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{218}{1995}\\x=\dfrac{244.}{665}\end{matrix}\right.\)

d) \(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=\dfrac{18}{19}-1\dfrac{2}{5}\)

\(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=-\dfrac{43}{95}\)

\(\left|\dfrac{1}{23}x\right|=-\dfrac{43}{95}-\dfrac{18}{90}\)

\(\left|\dfrac{1}{23}x\right|=-\dfrac{62}{95}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{23}x=\dfrac{62}{95}\\\dfrac{1}{23}x=-\dfrac{62}{95}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\dfrac{1}{95}\\x=-15\dfrac{1}{95}\end{matrix}\right.\)

Answer 3

a)

\(\dfrac{x}{27}=\dfrac{-3}{x}\\ \Leftrightarrow x^2=-81\\ \Leftrightarrow x\in\varnothing\)

b)

\(-\dfrac{9}{x}=\dfrac{-x}{\dfrac{4}{49}}\\ \Leftrightarrow-9.\dfrac{4}{49}=-x^2\\ \Leftrightarrow-x^2=\dfrac{-36}{49}\\ \Leftrightarrow x^2=\dfrac{36}{49}\\ \Leftrightarrow x=\pm\dfrac{6}{7}\)

c)

\(\left|7x-\dfrac{5}{3}\right|+\dfrac{7}{19}=-\dfrac{8}{15}\\ \Rightarrow\left|7x-\dfrac{5}{3}\right|=\dfrac{-257}{285}\\ \)

Mà \(\left|7x-\dfrac{5}{3}\right|\ge0\Rightarrow x\in\varnothing\)

d) \(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=\dfrac{18}{19}-1\dfrac{2}{5}\\ \Rightarrow\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=-\dfrac{43}{95}\\ \Rightarrow\left|\dfrac{1}{23}x\right|=-\dfrac{62}{95}\\ \Rightarrow x\in\varnothing\)

Answer 4

a) \(\Rightarrow\) xx = 27 \(\times\) (-3 ) \(\Rightarrow\) x² = -81 ( vô lí )

Vậy không tìm được x

b) \(\Rightarrow\) -xx = -9 \(\times\) 4/49 \(\Rightarrow\) -x² = -36/49

\(\Rightarrow\) -x² = - 6²/7² \(\Rightarrow\) x = 6/7