câu E

\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)

câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )

3, Tìm x, biết

\(d,\dfrac{-16}{x}=\dfrac{x}{-4}=>x^2=\left(-16\right).\left(-4\right)=>x^2=64\)

\(=>x=8\) hay \(x=-8\)

\(e,\dfrac{x}{-2}=\dfrac{\dfrac{8}{25}}{-x}=>-x^2=-2.\dfrac{8}{5}=\dfrac{-16}{25}\)

\(=>-x^2=0,64=>x=0,8\)

\(g,\dfrac{x}{-15}=\dfrac{-60}{x}\)

\(=>x^2=\left(-15\right).\left(-60\right)\)\(=>x^2=900=>x=30\) hay \(x=-30\)

d) \(\dfrac{-16}{x}=\dfrac{x}{-4}\)

= 16 . 4 = x.x

= 64 = \(x^2\)

= \(8^2=x^2\)

vậy x = 8

e)\(\dfrac{x}{-2}=\dfrac{8}{\dfrac{25}{-x}}\)

= -2 . \(\dfrac{8}{25}\) = -x . x

= -0,64 = \(-x^2\)

= 0,64 = \(x^2\)

0,8\(^2=x^2\)

vậy x = 0,8

g) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)

= -15 . -60 = x.x

= 900 = \(x^2\)

30 \(^2=x^2\)

vậy x = 30

\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\left(ĐK:x\ne0\right)\)

\(\Rightarrow\dfrac{2}{x}=\dfrac{x}{\dfrac{8}{25}}\)

\(\Rightarrow\dfrac{2}{x}=x\cdot\dfrac{25}{8}\)

\(\Rightarrow\dfrac{2}{x}=\dfrac{25x}{8}\)

\(\Rightarrow25x\cdot x=2\cdot8\)

\(\Rightarrow25x^2=16\)

\(\Rightarrow x^2=\dfrac{16}{25}\)

\(\Rightarrow x^2=\left(\dfrac{4}{5}\right)^2\)

\(\Rightarrow x=\pm\dfrac{4}{5}\)

Vậy:\(x\in\left\{-\dfrac{4}{5};\dfrac{4}{5}\right\}\)

-2/x = -x/(8/25)

x.(-x) = -2.(8/25)

-x² = -16/25

x² = 16/25

x = -4/5 hoặc x = 4/5

a) \(x^2=\left(-15\right).\left(-60\right)=900=>x=\)\(\pm\)\(30\)

b) \(-x^2=\dfrac{-16}{25}=>x^2=\dfrac{16}{25}=>x=\)\(\pm\)\(\dfrac{4}{5}\)

a)\(\dfrac{x}{-15}\)= \(-\dfrac{60}{x}\)

=> x . x = -15 . (-60)

=> \(^{x^2}\) = 900

x = 30

b) \(-\dfrac{2}{x}\) = \(-\dfrac{x}{\dfrac{8}{25}}\)

=> -2 . \(\dfrac{8}{25}\) = x . (-x)

=> \(\dfrac{-16}{25}\) = \(^{x^2}\)

=> x = \(\dfrac{4}{5}\)và \(-\dfrac{4}{5}\)

nhớ tích cho mk vs nha >_<

a: =>\(-\dfrac{6+x}{2}-\dfrac{3}{2}=2\)

=>-x-6-3=4

=>-x-9=4

=>-x=5

hay x=-5

b: =>(x+1)²=16

=>x+1=4 hoặc x+1=-4

=>x=3 hoặc x=-5

c: \(\Leftrightarrow\left(\dfrac{x-2}{27}-1\right)+\left(\dfrac{x-3}{26}-1\right)+\left(\dfrac{x-4}{25}-1\right)+\left(\dfrac{x-5}{24}-1\right)+\left(\dfrac{x-44}{5}+3\right)=0\)

=>x-29=0

hay x=29

bài 1) ta có : \(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow2\left(x+y\right)=3\left(2x-y\right)\)

\(\Leftrightarrow2x+2y=6x-3y\Leftrightarrow4x=5y\Leftrightarrow\dfrac{x}{y}=\dfrac{5}{4}\)

vậy \(\dfrac{x}{y}=\dfrac{5}{4}\)

bài 1

\(\dfrac{2x-y}{x+y}=\dfrac{2}{3}\Leftrightarrow\dfrac{2.\dfrac{x}{y}-1}{\dfrac{x}{y}+1}=\dfrac{2.\dfrac{x}{y}+2-3}{\dfrac{x}{y}+1}=2-\dfrac{3}{\dfrac{x}{y}+1}=\dfrac{2}{3}\)

\(2-\dfrac{2}{3}=\dfrac{4}{3}=\dfrac{3}{\dfrac{x}{y}+1}\)

\(\left(\dfrac{x}{y}+1\right)=\dfrac{9}{4}\Rightarrow\dfrac{x}{y}=\dfrac{9}{4}-\dfrac{4}{4}=\dfrac{5}{4}\)

Ơ mọe

Sao biết rồi còn gửi câu hỏi

hì bạn cứ trả lời đi

a: Đặt A=0

=>-2/3x=5/9

hay x=-5/6

b: Đặt B(x)=0

=>(x-2/5)(x+2/5)=0

=>x=2/5 hoặc x=-2/5

c: Đặt C(X)=0

\(\Leftrightarrow x^3\cdot\dfrac{1}{2}=-\dfrac{4}{27}\)

\(\Leftrightarrow x^3=-\dfrac{8}{27}\)

hay x=-2/3

Bài 2:

a: =>x^2=60

=>\(x=\pm2\sqrt{15}\)

b: =>2^2x+3=2^3x

=>3x=2x+3

=>x=3

c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)

\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)

=>1/2x-2=4

=>1/2x=6

=>x=12

Bài 1:

a) \(\dfrac{x}{15}=\dfrac{-2}{3,5}\)\(\Rightarrow x=\dfrac{15\cdot\left(-2\right)}{3,5}=-\dfrac{60}{7}\)

b) \(\dfrac{16}{x}=\dfrac{x}{25}\)\(\Rightarrow x^2=16\cdot25\Rightarrow x^2=400\Rightarrow x=\pm20\)

c) \(\dfrac{0,5}{0,7}=\dfrac{-0,1}{5x}\)\(\Rightarrow5x=\dfrac{\left(-0,1\right)\cdot0,7}{0,5}=-\dfrac{7}{50}\Rightarrow x=\dfrac{-\dfrac{7}{50}}{5}=-0,028\)

Bài 3:

a) Theo đề, ta có:

\(\dfrac{x}{5}=\dfrac{y}{25}\) và \(x+y=60\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{5}=\dfrac{y}{25}=\dfrac{x+y}{5+25}=\dfrac{60}{30}=2\)

\(\Rightarrow\dfrac{x}{5}=2\Rightarrow x=10\)

\(\Rightarrow\dfrac{y}{25}=2\Rightarrow y=50\)

b) Theo đề ta có:

\(5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\) và \(x-y=-5\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x-y}{3-5}=\dfrac{-5}{-2}=2,5\)

\(\Rightarrow\dfrac{x}{3}=2,5\Rightarrow x=7,5\)

\(\Rightarrow\dfrac{y}{5}=2,5\Rightarrow y=12,5\)

c) Theo đề ta có:

\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\) và \(y+z-x=8\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}=\dfrac{y+z-x}{4+6-2}=\dfrac{8}{8}=1\)

\(\Rightarrow\dfrac{x}{2}=1\Rightarrow x=2\)

\(\Rightarrow\dfrac{y}{4}=1\Rightarrow y=4\)

\(\Rightarrow\dfrac{z}{6}=1\Rightarrow z=6\)

d) Theo đề ta có

\(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}\left(1\right)\)

\(\dfrac{y}{6}=\dfrac{z}{8}\Rightarrow\dfrac{y}{12}=\dfrac{z}{16}\left(2\right)\)

Từ (1) và (2)\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}\) và \(x+y-z=50\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{16}=\dfrac{x+y-z}{9+12-16}=\dfrac{50}{5}=10\)

\(\Rightarrow\dfrac{x}{9}=10\Rightarrow x=90\)

\(\Rightarrow\dfrac{y}{12}=10\Rightarrow y=120\)

\(\Rightarrow\dfrac{z}{16}=10\Rightarrow z=160\)

e) Theo đề ta có:

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}\)và \(2x+3y+5z=86\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{5}=\dfrac{2x+3y+5z}{2\cdot3+3\cdot4+5\cdot5}=\dfrac{86}{43}=2\)

\(\Rightarrow\dfrac{x}{3}=2\Rightarrow x=6\)

\(\Rightarrow\dfrac{y}{4}=2\Rightarrow y=8\)

\(\Rightarrow\dfrac{z}{5}=2\Rightarrow z=10\)

f) Theo đề ta có

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}\)và \(x+y+z=-28\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{x+y+z}{2+5+7}=\dfrac{-28}{14}=-2\)

\(\Rightarrow\dfrac{x}{2}=-2\Rightarrow x=-4\)

\(\Rightarrow\dfrac{y}{5}=-2\Rightarrow y=-10\)

\(\Rightarrow\dfrac{z}{7}=-2\Rightarrow z=-14\)

g) Theo đề ta có

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{2}\) và \(2x^2+y^2+3z^2=316\)

Áp dụng tính chất dãy tỉ số bằng nhau

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{z}{2}=\dfrac{2x^2+y^2+3z^2}{2\cdot3^2+7^2+3\cdot2^2}=\dfrac{316}{79}=4\)

\(\Rightarrow\dfrac{x}{3}=4\Rightarrow x=12\)

\(\Rightarrow\dfrac{y}{7}=4\Rightarrow y=28\)

\(\Rightarrow\dfrac{z}{2}=4\Rightarrow z=8\)