a) Ta có :  2 x : 2 2 = 2 5  nên x = 7.

b) Ta có:  3 x : 3 2 = 3 5  nên x = 7.

c) Ta có :  4 4 : 4 x = 4 2  nên x = 2.

d) Ta có :  5 x : 5 2 = 5 2  nên x = 4,

e) Ta có:  5 x + 1 : 5 = 5 4  nên x = 4.

f) Ta có :  4 2 x - 1 : 4 = 4 2  nên x = 2

Đề bài : Tìm Min của \(D=\sqrt{49x^2-42x+9}+\sqrt{49x^2+42x+9}\)

Ta có ; \(D=\sqrt{49x^2-42x+9}+\sqrt{49x^2+42x+9}=\sqrt{49\left(x-\frac{3}{7}\right)^2}+\sqrt{49\left(x+\frac{3}{7}\right)^2}=7\left(\left|x-\frac{3}{7}\right|+\left|x+\frac{3}{7}\right|\right)\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\). Dấu "=" xảy ra khi a,b cùng dấu.

Được; \(D=7\left(\left|\frac{3}{7}-x\right|+\left|x+\frac{3}{7}\right|\right)\ge7.\left|\frac{3}{7}-x+x+\frac{3}{7}\right|=7.\frac{6}{7}=6\)

\(\Rightarrow D\ge6\). Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+\frac{3}{7}\ge0\\\frac{3}{7}-x\ge0\end{cases}\Leftrightarrow}\frac{-3}{7}\le x\le\frac{3}{7}\)

Vậy Min D = 6 \(\Leftrightarrow\frac{-3}{7}\le x\le\frac{3}{7}\)

Mình thấy đề bài hơi kì kì ^^

Ta có ; \(D=2\sqrt{49x^2-42x+9}=2\sqrt{49\left(x-\frac{3}{7}\right)^2}=14\left|x-\frac{3}{7}\right|\ge0\)

Do đó Min D = 0 \(\Leftrightarrow x=\frac{3}{7}\)

viết lại đề bài: \(\sqrt{49x^2-42x+9}+\sqrt{49x^2+42x+9}\)

\(B=l7x-3l+l7x+3l\)

     = \(l3-7xl+l7x+3l\) \(\ge l3-7x+7x+3l=6\)

Vậy GTNN là 6 khi -7/3 <= x <= 7/3 

Ngu Người **** cho tui đi 

\(A=\sqrt{\left(7x-3\right)^2}+\sqrt{\left(7x+3\right)^2}\)

\(A=\left|7x-3\right|+\left|7x+3\right|=\left|3-7x\right|+\left|7x+3\right|\)

\(A\ge\left|3-7x+7x+3\right|=6\)

\(A_{min}=6\) khi \(\left(3-7x\right)\left(7x+3\right)\ge0\Rightarrow-\frac{3}{7}\le x\le\frac{3}{7}\)

4^2x+1+4^2x=20

4^2x.4^1+4^2x.1=20

4^2x.(4^1+1)=20

4^2x.5=20

4^2x=20:5

4^2x=4

4^2x=2^2

2x=2

x=2:2

x=1

sai đừng trách mk còn đúng thì 1 like

4^2x+1+4^2x=20

4^2x.4^1+4^2x.1=20

4^2x.(4^1+1)=20

4^2x.5=20

4^2x=20:5

4^2x=4

4^2x=2^2

2x=2

\(a,\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

\(=|2x-1|+|2x-3|\)

\(b,\sqrt{49x^2-42x+9}+\sqrt{49x^2+42x+9}\)

\(=\sqrt{\left(7x-3\right)^2}+\sqrt{\left(7x+3\right)^2}\)

\(=|7x-3|+|7x+3|\)

=.= hok tốt!!

\(\left(x^2-3x\right)^2-14x^2+42x+40=\left[\left(x^2-3x\right)^2-14\left(x^2-3x\right)+49\right]-9=\left(x^2-3x-7\right)-3^3=\left(x^2-3x-7-3\right)\left(x^2-3x-7+3\right)=\left(x^2-3x-10\right)\left(x^2-3x-4\right)=\left(x-5\right)\left(x+2\right)\left(x-4\right)\left(x+1\right)\)

\(\left(x^2-3x\right)^2-14x^2+42x+40\\ =x^4-6x^3+9x^2-14x^2+42x+40\\ =x^4-6x^3-5x^2+42x+40\\ =x^4+x^3-7x^3-7x^2+2x^2+2x+40x+40\\ =\left(x+1\right)\left(x^3-7x^2+2x+40\right)\\ =\left(x+1\right)\left(x^3+2x^2-9x^2-18x+20x+40\right)\\ =\left(x+1\right)\left(x+2\right)\left(x^2-9x+20\right)\\ =\left(x-5\right)\left(x-4\right)\left(x+1\right)\left(x+2\right)\)

\(\left(x^2-3x\right)^2-14x^2+42x+40\)

\(=\left(x^2-3x\right)^2-14\left(x^2-3x\right)+40\)

\(=\left(x^2-3x-4\right)\left(x^2-3x-10\right)\)

\(=\left(x-4\right)\left(x+1\right)\left(x-5\right)\left(x+2\right)\)

\(\Leftrightarrow3\cdot4^{2x+1}=36+156=192\)

=>2x+1=3

hay x=1

\(3.4^{2x+1}-156=72:2\)

\(3.4^{2x+1}-156=36\)

\(4^{2x+1}-156=36:3\)

\(4^{2x+1}-156=12\)

\(4^{2x+1}=12+156\)

\(4^{2x+1}=168\)

\(2x+1=168:4\)

\(2x+1=42\)

\(x+1=42:2\)

\(x+1=21\)

\(x=21-1\)

\(x=20\)

Câu 1: 

\(\sqrt{x^2-2x+1}+\sqrt{x^2-4x+4}=3\)

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

Trường hợp 1: x<1

(1) trở thành 1-x+2-x=3

=>3-2x=3

=>x=0(nhận)

Trường hợp 2: 1<=x<2

(1) trở thành x-1+2-x=3

=>1=3(loại)

Trường hợp 3: x>=2

(1) trở thành x-1+x-2=3

=>2x-3=3

=>2x=6

hay x=3(nhận)

a) P=\(\sqrt{4x^2-4x+1}+\sqrt{4x^2-12x+9}=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)

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

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

<=> \(P\ge2\)

Dấu "=" xảy ra <=> (2x-1)(3-2x)\(\ge0\)

<=> \(\frac{1}{2}\le x\le\frac{3}{2}\)

Vậy min P=2 <=>\(\frac{1}{2}\le x\le\frac{3}{2}\)

b)Tương tự ý a