a, \(x^2=\dfrac{1}{16}\Rightarrow x=\pm\dfrac{1}{4}\)

b, \(x^5:x^2=-\dfrac{1}{64}\Rightarrow x^3=\left(-\dfrac{1}{4}\right)^3\Rightarrow x=-\dfrac{1}{4}\)

c, \(x^3:x^2=\dfrac{32}{243}\Rightarrow x=\dfrac{32}{243}\)

d, \(\left(x^2\right)^2=\dfrac{81}{16}\Rightarrow x^4=\left(\dfrac{3}{2}\right)^4\Rightarrow x=\pm\dfrac{3}{2}\)

Chúc bạn học tốt!!!

3) Tìm x

a) \(^{x^2}\)=\(\dfrac{1}{16}\)

<=> x = \(\sqrt{-\dfrac{1}{16}}\)

\(\sqrt{\dfrac{1}{16}}\)

<=> x = -14

+14

b) \(x^{5^{ }}\): \(x^2\) = \(-\dfrac{1}{64}\)

<=> \(^{x^{5-2}}\) =\(-\dfrac{1}{64}\)

<=> \(x^3\) = \(-\dfrac{1}{64}\)

<=> x = \(-\dfrac{1}{4}\)

c)\(x^3:x^2\) = \(\dfrac{32}{243}\)

<=> \(^{x^{3-2}}\) = \(\dfrac{32}{243}\)

<=> x = \(\dfrac{32}{243}\)

d) \((x^2)^2\) = \(\dfrac{81}{16}\)

<=>\(^{x^{2.2}}\) = \(\dfrac{81}{16}\)

<=> \(x^4\) = \(\dfrac{81}{16}\)

<=> x = \(\dfrac{3}{2}\)

\(-\dfrac{3}{2}\)

a) Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=7.2=14\\y=13.2=26\end{matrix}\right.\)

Vật \(x=14;y=26\)

b) (Chỗ này bạn viết nhầm thì phải)

Ta có:

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

và \(x-y=-16\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{-16}{-4}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.4=12\\y=7.4=28\end{matrix}\right.\)

Vậy \(x=12;y=28\)

c) Ta có:

\(\dfrac{x}{19}=\dfrac{y}{21}=\dfrac{2x}{38}\)

và \(2x-y=34\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)

\(\Rightarrow\left\{{}\begin{matrix}2x=38.2=76\Rightarrow x=38\\y=21.2=42\end{matrix}\right.\)

Vậy \(x=38;y=42\)

d) Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=9.4=36=6^2=\left(-6\right)^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\\y^2=16.4=64=8^2=\left(-8\right)^2\Rightarrow\left[{}\begin{matrix}y=8\\y=-8\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x:y\right)\in\left\{\left(6;8\right);\left(6;-8\right);\left(-6;8\right);\left(-6;-8\right)\right\}\)

Cả 4 cái có 1 câu huyền thoại:"Áp dụng tính chất dãy tỉ số = nhau ta có" nên mk nói cho cả 4 lun :v

a) \(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)

b) \(\dfrac{x}{19}=\dfrac{y}{21}\Rightarrow\dfrac{2x}{38}=\dfrac{y}{21}=\dfrac{2x-y}{38-21}=\dfrac{34}{17}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.19=38\\y=2.21=42\end{matrix}\right.\)

c) \(7x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{-16}{-4}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.3=12\\y=4.7=28\end{matrix}\right.\)

c) \(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4.9=36\Rightarrow x=\pm6\\y^2=4.16=64\Rightarrow y=\pm8\end{matrix}\right.\)

\(\)

câu a) mình chịu (dùng kiến thức lớp 12 chắc làm đc haha)

b) gt ⇒ \(\frac{1}{6}.6^{x+2}-6^x=6^{14}-6^{13}\)

⇒ \(6^{x+1}-6^x=6^{14}-6^{13}\)

⇒ \(6^x\left(6-1\right)=6^{13}\left(6-1\right)\)

⇒ \(x=13\)

c) gt ⇒ \(\frac{1}{2}.2^{x+4}-2^x=2^{13}-2^{10}\)

⇒ \(2^{x+3}-2^x=2^{13}-2^{10}\)

⇒ \(2^x\left(2^3-1\right)=2^{10}\left(2^3-1\right)\)

⇒ \(x=10\)

d) gt ⇒ \(\frac{1}{3}.3^{x+4}-4.3^x=3^{16}-4.3^{13}\)

⇒ \(3^{x+3}-4.3^x=3^{16}-4.3^{13}\)

⇒ \(3^x\left(3^3-4\right)=3^{13}\left(3^3-4\right)\)

⇒ \(x=13\)

câu d chưa có đóng ngoặc kìa bn

Bài 1:

a) \(x^2-3=1\)

\(\Rightarrow x^2=1+3=4\)

\(\Rightarrow x=\pm2\)

b)\(2x^3+12=-4\)

\(\Rightarrow2x^3=-4-12=-16\)

\(\Rightarrow x^3=-8\)

\(\Rightarrow x=-2\)

c)\(\left(2x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=\dfrac{7}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)

a) \(x^2-3=1\Rightarrow x^2=4\Rightarrow x=\pm2\)

b) \(2x^3+12=-4\Rightarrow2x^3=-16\)

\(\Rightarrow x^3=-\dfrac{16}{2}=-8=-2^3\)

\(\Rightarrow x=-2\)

c) \(\left(2x-3\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=4\\2x-3=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d,h,i,k cững tương tự....

2. Tham khảo thêm tại đây nha bạn

https://hoc24.vn/hoi-dap/question/417550.html

\(\dfrac{x^2}{9}=\dfrac{y^2}{16}\&x^2+y^2=100\)

Áp dụng tc dãy tỉ số bằng nhau ta được:

\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{9}=4\\\dfrac{y^2}{16}=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=6\\y=8\end{matrix}\right.\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{x^2+y^2}{9+16}=\dfrac{100}{25}=4\)

Do đó: \(x^2=36;y^2=64\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{6;-6\right\}\\y\in\left\{8;-8\right\}\end{matrix}\right.\)

a: =>13/6x=-1/2

=>x=-1/2:13/6=-1/2x6/13=-6/26=-3/13

b: =>2x-1=1/2 hoặc 2x-1=-1/2

=>2x=3/2 hoặc 2x=1/2

=>x=3/4 hoặc x=1/4

c: =>(x-4)(x+4)(4-5x)=0

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

1: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^6\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{18}\)

=>4x=18

hay x=9/2

2: \(\left(\dfrac{1}{16}\right)^x=\left(\dfrac{1}{8}\right)^{36}\)

\(\Leftrightarrow\left(\dfrac{1}{2}\right)^{4x}=\left(\dfrac{1}{2}\right)^{108}\)

=>4x=108

hay x=27

3: \(\left(\dfrac{1}{81}\right)^x=\left(\dfrac{1}{27}\right)^4\)

\(\Leftrightarrow\left(\dfrac{1}{3}\right)^{4x}=\left(\dfrac{1}{3}\right)^{12}\)

=>4x=12

hay x=3

a, ta co:

x-2/4=-16/2-x

=>(x-2)(2-x)=(-16).4

lai co: x-2/2-x=-1

=>x-2=(-1).(2-x)

a, ta co:

x-2/4=-16/2-x

=>(x-2)(2-x)=(-16).4 (1)

lai co: x-2/2-x=-1

=>x-2=(-1).(2-x) (2)

thay(2) vao(1) ,ta co:

(2-x)^2=-64

.........(tu lam tiep nha)

Ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}\Leftrightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{3}\right)^2\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{x^2-y^2}{4-9}=\dfrac{-16}{-5}=3,2\)

Do đó:

x² = \(4.3,2=12,8\Rightarrow x=\sqrt{12,8}\)

y² = \(9.3,2=28,8\Rightarrow y=\sqrt{28,8}\)

Suy ra: \(\dfrac{\sqrt{28,8}}{4}=\dfrac{z}{5}\Rightarrow z=\dfrac{5.\sqrt{28,8}}{4}=3\sqrt{5}\)

Vậy \(x=\sqrt{12,8};y=\sqrt{28,8};z=3\sqrt{5}\)