0,4 :x = x: 0.9

=> x.x = 0,4 . 0,9

=> x² = 0,36

=> x= 0,6 hoặc x= -0,6

vậy x=  0,6 hoặc x= -0,6

\(\frac{0.4}{x}=\frac{x}{0.9}=>x^2=0.4\times0.9=\frac{9}{25}\)

\(=>x=\frac{3}{5}\)

0,4 : x = x : 0,9

=> 0,4 . 0,9 = x . x

=> 0,36 = x²

=> (0,6)² = x²

=> x = 0,6

0,4 : x = x : 0,9

= > x.x = 0,4.0,9

= > x² = 0,36

= > x = 0,6 hoặc x = -0,6

Vậy x = 0,6 hay x = -0,6

\(0,4:x=x:0,9\)

\(\Rightarrow x^2=0,4.0,9=0,36\)

=>\(x=\pm0,6\)

\(0,4:x=x:0,9\)

\(\Rightarrow x^2=0,4.0,9\)

\(\Rightarrow x^2=0,36\)

\(\Rightarrow x=\sqrt[2]{0,36}\)

\(\Rightarrow x=0,6\)

\(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\dfrac{x}{y}=\dfrac{3}{7}\)

\(\dfrac{x}{y}-1=\dfrac{-5}{19}\Rightarrow\dfrac{x}{y}=\dfrac{14}{19}\)

Vô lí => không có x,y thỏa mãn

a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{7}\)

nên \(\dfrac{x}{y}=\dfrac{3}{7}\)

b) Ta có: \(\dfrac{x}{y-1}=\dfrac{5}{-19}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y-1}{-19}\)

hay \(\dfrac{x}{5}=\dfrac{1-y}{19}\)

0,4 : x = x : 0,9 

(=) x² = 0,4 x 0,9 

(=) x² = 0,36

(=) x = \(\sqrt{0,36}\)

(=) x = { -6 ; 6 }

nha ban

\(\frac{0,4}{x}=\frac{x}{0,9}\Rightarrow x^2=0,4\times0,9=0,36\Rightarrow x=0,6\)hoặc x = -0,6

Vậy x = {0,6 ; -0,6}

0,4:x=x:0.9

\(\Rightarrow\frac{0,4}{x}=\frac{x}{0,9}\Leftrightarrow x.x=0,4.0,9\Leftrightarrow x^2=0,36\)

ma x²=0,6²

=>x=0,6 hoac -0,6

a)4/3:x=-4/7                         b)5/3.x=7/12

⇒x=-16/21                             ⇒x=7/20

a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\)         ĐKXĐ: x \(\ne\) 0

<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)

<=> 15x + 28 = 3x

<=> 15x - 3x = -28

<=> 12x = -28

<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)

b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)

<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)

<=> -5x . 6 = 12 . 2

<=> -30x = 24

<=> x = \(-\dfrac{4}{5}\)

Giải:

a) \(\left(x-1\right)\left(y+2\right)=7\) 

\(\Rightarrow\left(x-1\right)\) và \(\left(y+2\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\) 

Ta có bảng giá trị:

Vậy \(\left(x;y\right)=\left\{\left(-6;-3\right);\left(0;-9\right);\left(2;5\right);\left(8;-1\right)\right\}\) 

b) \(\left(x-2\right)\left(3y+1\right)=17\) 

\(\Rightarrow\left(x-2\right)\) và \(\left(3y+1\right)\inƯ\left(17\right)=\left\{\pm1;\pm17\right\}\) 

Ta có bảng giá trị:

Vậy \(\left(x;y\right)=\left\{\left(1;-6\right);\left(19;0\right)\right\}\)

Ko ghi lại đề nhé 

a) \(TH1\left[{}\begin{matrix}x-1=1\\y+2=7\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\y=5\end{matrix}\right.\)

\(TH2:\left[{}\begin{matrix}x-1=-1\\y+2=-7\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\y=-9\end{matrix}\right.\)

\(TH3:\left[{}\begin{matrix}x-1=7\\y+2=1\end{matrix}\right.=>\left[{}\begin{matrix}x=8\\y=-1\end{matrix}\right.\)

\(TH4:\left[{}\begin{matrix}x-1=-7\\y+2=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-6\\y=-3\end{matrix}\right.\)

b) \(TH1:\left[{}\begin{matrix}x-2=1\\3y+1=17\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\y=\dfrac{16}{3}\end{matrix}\right.=>Loại\)

\(TH2:\left[{}\begin{matrix}x-2=-1\\3y+1=-17\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\y=-6\end{matrix}\right.Chọn\)

\(TH3:\left[{}\begin{matrix}x-2=17\\3y+1=1\end{matrix}\right.=>\left[{}\begin{matrix}x=19\\y=0\end{matrix}\right.=>Chọn\)

\(TH4:\left[{}\begin{matrix}x-2=-17\\3y+1=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=-15\\y=\dfrac{-2}{3}\end{matrix}\right.=>Loại\)

Bạn tự kết luận hộ mk nha

a: \(x-43=\left(35-x\right)-48\)

=>\(x-43=35-x-48\)

=>\(x-43=-x-13\)

=>\(x+x=-13+43\)

=>2x=30

=>x=30/2=15

b: \(305-x+14=48+\left(x-23\right)\)

=>\(319-x=48+x-23=25+x\)

=>\(x+25=319-x\)

=>\(x+x=319-25\)

=>\(2x=294\)

=>\(x=\dfrac{294}{2}=147\)

c: \(-\left(-x-6+85\right)=\left(x+51\right)-54\)

=>\(-\left(-x+79\right)=x+51-54\)

=>x-79=x-3

=>-79=-3(vô lý)

=>\(x\in\varnothing\)

d: \(-\left(35-x\right)-\left(37-x\right)=33-x\)

=>\(-35+x-37+x=33-x\)

=>2x-72=-x+33

=>\(2x+x=33+72\)

=>3x=105

=>\(x=\dfrac{105}{3}=35\)

Giải:

\(0,4:x=x:0,9\)

\(\Leftrightarrow\dfrac{0,4}{x}=\dfrac{x}{0,9}\)

\(\Leftrightarrow x.x=0,4.0,9\)

\(\Leftrightarrow x^2=0,36\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0,6\\x=-0,6\end{matrix}\right.\)

Vậy ...

\(\dfrac{2}{\dfrac{5}{x}}=\dfrac{x}{\dfrac{9}{10}}\)\(\Rightarrow\)x^2=\(\dfrac{2}{5}.\dfrac{9}{10}=\dfrac{9}{25}\)

Nên x=\(\dfrac{3}{5}\)hoặc x=\(\dfrac{-3}{5}\)