\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)

 TH1:   \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)

TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)

\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)

\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)

\(\Rightarrow x=\frac{2}{5}\)

\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)

\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)

\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)

\(\Rightarrow3x=\frac{1}{9}\)

\(\Rightarrow x=\frac{1}{27}\)

\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)

\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\)  \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)

a)\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)

=\(\left(x+\frac{1}{5}\right)^2=\frac{9}{25}=\frac{3^2}{5^2}\)

=\(x+\frac{1}{5}=\frac{3}{5}\)

\(x=\frac{2}{5}\)

b)\(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=\frac{24}{27}\)

=\(x=-\frac{35}{27}\)

a) \(x=\frac{1}{27}\)

b) \(x=\frac{2}{5}\)

ủng hộ mk nha bn

bạn trả lời giúp

3) \(\left(x+\dfrac{1}{5}\right)^2\) + \(\dfrac{17}{25}\) = \(\dfrac{26}{25}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{26}{25}\) - \(\dfrac{17}{25}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{9}{25}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\dfrac{3}{5}.\dfrac{3}{5}\)

=> \(\left(x+\dfrac{1}{5}\right)^2\) = \(\left(\dfrac{3}{5}\right)^2\)

=> \(x\) + \(\dfrac{1}{5}\) = \(\dfrac{3}{5}\)

=> \(x\) = \(\dfrac{3}{5}\) - \(\dfrac{1}{5}\)

=> \(x\) = \(\dfrac{2}{5}\)

4) -1\(\dfrac{5}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-24}{27}\)

=> \(\dfrac{-32}{27}\) - \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{9}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-32}{27}\) - \(\dfrac{-8}{9}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-8}{27}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\) . \(\dfrac{-2}{3}\)

=> \(\left(3x-\dfrac{7}{9}\right)^3\) = \(\left(\dfrac{-2}{3}\right)^3\)

=> \(3x-\dfrac{7}{9}=\dfrac{-2}{3}\)

=> \(3x=\dfrac{-2}{3}+\dfrac{7}{9}\)

=> \(3x=\dfrac{1}{9}\)

=> \(x=\dfrac{1}{9}:3\)

=> \(x=\dfrac{1}{27}\)

a, <=>(-5)^x=(-5)³

<=> x=3

b, <=> 5^2x=5²²

<=> x=11

c, 3^2x=3¹⁷.3¹⁵

<=> 3^2x=3³²

<=> x=16

d,2^x+1=2²⁵

<=> x+1=25

<=> x=24

Chúc hok tốt!!!

a)x=3

b)x=11

c)x=16

d)x=24

a,(5/8/17+-4/17):x+33/182=4/11

=5/4/17:x+33/182=4/11

5/4/17:x=4/11-33/182

5/4/17:x=365/2002

x=5/4/17:365/2002

x=28/4438/6205

b,-1/5/27-(3x-7/9)^3=-24/27

(3x-7/9)^3=-1/5/27--24/27

(3x-7/9)^3=-8/27

(3x-7/9)^3=(-2/3)^3

3x-7/9=-2/3

3x=-2/3+7/9

3x=1/9

x=1/9:3

x=1/27

1/27 k nha!

Bài 2:

a) \(\left(x-3\right)^3+27=0\)

\(\Leftrightarrow\left(x-3\right)^3=0-27\)

\(\Leftrightarrow\left(x-3\right)^3=-27\)

\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)

\(\Leftrightarrow x-3=-3\)

\(\Leftrightarrow x=\left(-3\right)+3\)

\(\Leftrightarrow x=0\)

b) \(-125-\left(x+1\right)^3=0\)

\(\Leftrightarrow\left(x+1\right)^3=-125-0\)

\(\Leftrightarrow\left(x+1\right)^3=-125\)

\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)

\(\Leftrightarrow x+1=-5\)

\(\Leftrightarrow x=\left(-5\right)-1\)

\(\Leftrightarrow x=-6\)

c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

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

\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)

\(\Leftrightarrow2x=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}:2\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

d) \(2^x+2^{x+1}=24\)

\(\Leftrightarrow2^x+2^x.2=24\)

\(\Leftrightarrow2^x\left(1+2\right)=24\)

\(\Leftrightarrow2^x.3=24\)

\(\Leftrightarrow2^x=24:3\)

\(\Leftrightarrow2^x=8\)

\(\Leftrightarrow2^x=2^3\)

\(\Rightarrow x=3\)

e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)

\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)

g) \(\left|x-3\right|+2x=10\)

\(\Leftrightarrow\left|x-3\right|=10-2x\)

\(\Leftrightarrow\left|x-3\right|=2.5-2x\)

\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)

(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)

Bài 1:

a) \(2^7+2^9⋮10\)

Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)

\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)

\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)

\(\Leftrightarrow\overline{C8}+\overline{D2}\)

\(\Leftrightarrow\overline{E0}\)

Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)

b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)

Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)

\(\Leftrightarrow2^{72}.5^{20}\)

Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)

c) \(3^{10}+3^{12}⋮30\)

Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)

\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)

\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)

\(\Leftrightarrow\overline{C9}+\overline{B1}\)

\(\Leftrightarrow\overline{D0}⋮10\)

(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)