Ta có: \(f\left(x\right)-g\left(x\right)=\left(0,7x^4+0,2x^2-5\right)-\left(0,3x^4+0,1x^2-8\right)\)

\(=0,7x^4+0,2x^2-5-0,3x^4-0,1x^2+8\)

\(=0,4x^4+0,1x^2+3\)

Vì \(\hept{\begin{cases}0,4x^4\ge0\\0,1x^2\ge0\end{cases}}\)nên \(0,4x^4+0,1x^2+3>0\)

\(\Rightarrow f\left(x\right)-g\left(x\right)>0\)hay \(f\left(x\right)>g\left(x\right)\forall x\)

Ta có :

\(\left|0,2x-3,1\right|+\left|0,2+3,1\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left|0,2x-3,1\right|=0\\\left|0,2x+3,1\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0,2x-3,1=0\\0,2x+3,1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}0,2x=3,1\\0,2x=-3,1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=15,5\\x=-15,5\end{matrix}\right.\) (vô lí)

Vậy ko tìm dc x thỏa mãn theo yêu cầu

Ta có: \(\left|0.2x-3.1\right|\)\(\ge\)0 vs mọi x

\(\left|0.2x+3.1\right|\)\(\ge\)0 vs mọi x

nên \(\left|0.2x-3.1\right|+\left|0.2x+3.1\right|=0\)

<=> \(\left\{{}\begin{matrix}\left|0.2x-3.1\right|=0\\\left|0.2x+3.1\right|=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}0.2x-3.1=0\\0.2x+3.1=0\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}0.2x=3.1\\0.2x=-3.1\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}x=15.5\\x=-15.5\end{matrix}\right.\) (vô lý, vì 1 giá trị x ko thể có 2 giá trị)

Vậy ko có giá trị x nào thỏa mãn \(\left|0.2x-3.1\right|+\left|0.2x+3.1\right|=0\)

sao bạn lại phải cầu xin

1) MK không chép lại đề nữa nhé!

3^x = (3² ) ^-6 . (3³ ) ^-5 . (3⁴)⁸

3 ^x = 3^-12.3^-15 . 3³²

3^x= 3^-27.3³²

3^x= 3⁵

=>x= 5

Vậy x=5

\(\dfrac{7}{3}:0,2x=\dfrac{7}{6}:0,8\)

⇔ \(\dfrac{7}{3}:0,2x=\dfrac{35}{24}\)

⇔ \(0,2x=\dfrac{7}{3}:\dfrac{35}{24}\)

⇔ \(0,2x=1,6\)

⇔ x = 1,6 : 0,2

⇔ x = 8

Vậy x =8

Bài 1 :

Áp dụng t.c tỉ lệ thức là làm dc thôi bn

Bài 2 :

Đặt :

\(\dfrac{x}{4}=\dfrac{y}{7}=k\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4k\\y=7k\end{matrix}\right.\)\(\left(1\right)\)

Thay \(xy=112\) vào \(\left(1\right)\) ta có :

\(4k.7k=112\)

\(\Leftrightarrow28k^2=112\)

\(\Leftrightarrow k^2=4\)

\(\Leftrightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

+) \(k=2\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4.2=8\\y=7.2=14\end{matrix}\right.\)

+) \(k=-2\)\(\Leftrightarrow\left\{{}\begin{matrix}x=4.\left(-2\right)=-8\\y=7.\left(-2\right)=-14\end{matrix}\right.\)

Vậy ..

Ta có:

5^x+5^x-2=650\(\Rightarrow\)5^x.(1+5²)=650\(\Rightarrow\)5^x.26=650\(\Rightarrow\)5^x=25\(\Rightarrow\)x=2

2.5^2x-1-1=249\(\Rightarrow\)2.5^2x-1=250\(\Rightarrow\)5^2x-1=125=5³\(\Rightarrow\)2x-1=3\(\Rightarrow\)x=2

3.8^x-1=48\(\Rightarrow\)8^x-1=16(loại)

16^x-5-5=251\(\Rightarrow\)16^x-5=256=16²\(\Rightarrow\)x-5=2\(\Rightarrow\)x=7

Ta có : 5^x - 5^{x - 2} = 650

=> 5^x(1 + 5²) = 650

=> 5^x . 26 = 650

=> 5^x = 25

=> 5^x = 5² 

=> x = 2