sửa nha chỗ 24/7 thay = \(-\frac{24}{7}\)

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

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

\(\left(3x-\frac{7}{9}\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)

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

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

\(3x=\frac{1}{9}\)

\(x=\frac{1}{9}:3\)

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

(3x-7/9)³ = -32/27-(-24/27)

(3x-7/9)³ = -8/27

(3x)³-(7/9)³ = -8/27

(3x³)-343/729 = -8/27

(3x)³ = -8/27+ -343/729

(3x)³ = -559/729

3³ *x³ = -559/729

9*x³ = -559/729

x³ = -559/729: 9

x³ = -559/6561

x = -0,4400282812

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

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

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

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

\(\left(3x-\frac{7}{9}\right)=\sqrt[3]{-\frac{8}{27}}\)

Rồi làm tiếp đi

ủng hộ mk nha bn

bạn trả lời giúp

tk cho mk nhé mình đang bị âm nek

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

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

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

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

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

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

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

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

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

\(3x=\frac{1}{9}\)

\(x=\frac{1}{9}:3\)

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

\(V\text{ậy}\) \(x=\frac{1}{27}\)

Đây là cách làm đúng. Nhớ Lick cho mình nha !

Tô Phương Linh

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

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

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

\(\Leftrightarrow3x-\frac{7}{9}=-\frac{2}{3}\)

\(\Leftrightarrow3x=\frac{1}{9}\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Bổ sung câu a: \(\Rightarrow\) \(\left[\begin{array}{nghiempt}\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\\\left(x+\frac{1}{5}\right)^2=\left(-\frac{3}{5}\right)^2\end{array}\right.\)\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\) \(\Rightarrow\)  \(\left[\begin{array}{nghiempt}x=\frac{2}{5}\\x=-\frac{4}{5}\end{array}\right.\)

a) \(\frac{x}{5}-\frac{x}{6}=\frac{3}{10}\\ \frac{6x}{30}-\frac{5x}{30}=\frac{3\cdot3}{10\cdot3}\\ \frac{x}{30}=\frac{9}{30}\\ \Rightarrow x=9\) Vậy x = 9

b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\\ \frac{-32}{27}+\frac{24}{27}=\left(3x-\frac{7}{9}\right)^3\\ \left(3x-\frac{7}{9}\right)^3=\frac{-8}{27}\\ \left(3x-\frac{7}{9}\right)^3=\left(\frac{-2}{3}\right)^3\\ \Rightarrow3x-\frac{7}{9}=\frac{-2}{3}\\ 3x=\frac{-2}{3}+\frac{7}{9}\\ 3x=\frac{-6}{9}+\frac{7}{9}\\ 3x=\frac{1}{9}\\ x=\frac{1}{9}:3\\ x=\frac{1}{9\cdot3}\\ x=\frac{1}{27}\)Vậy \(x=\frac{1}{27}\)

a) \(x=\frac{1}{27}\)

b) \(x=\frac{2}{5}\)

Bài làm:

c) \(\left(x-2\right)\left(x+3\right)>0\)

Ta xét 2 trường hợp sau:

+ Nếu \(\hept{\begin{cases}x-2>0\\x+3>0\end{cases}\Rightarrow}\hept{\begin{cases}x>2\\x>-3\end{cases}\Rightarrow}x>2\)

+ Nếu \(\hept{\begin{cases}x-2< 0\\x+3< 0\end{cases}}\Rightarrow\hept{\begin{cases}x< 2\\x< -3\end{cases}}\Rightarrow x< -3\)

Vậy \(\orbr{\begin{cases}x>2\\x< -3\end{cases}}\)

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

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

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

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

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

\(\Leftrightarrow3x=\frac{1}{9}\)

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

Vậy \(x=\frac{1}{27}\)

Học tốt!!!!