Ta có \(\frac{2}{5}x+\frac{1}{3}x=\frac{3}{5}\)

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

\(\Rightarrow x.\frac{8}{15}=\frac{3}{5}\)

\(\Rightarrow x=\frac{9}{8}\)

\(\frac{2}{5}\text{ x }x+\frac{1}{3}\text{ x }x=\frac{3}{5}\)

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

\(\frac{11}{15}\text{ x }x=\frac{3}{5}\)

\(x=\frac{3}{5}\text{ : }\frac{11}{15}\)

\(x=\frac{9}{11}\)

Với \(x\ge\dfrac{5}{2}\)có: \(A=x+\sqrt{2x-5}\ge\dfrac{5}{2}+0=\dfrac{5}{2}\)

Dấu '=' xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)

\(\Rightarrow A_{min}=\dfrac{5}{2}\)

\(\left(x-\frac{3}{5}\right)=\frac{2}{5}×-\frac{1}{3}\)

\(\left(x-\frac{3}{5}\right)=-\frac{2}{165}\)

\(x=-\frac{2}{165}+\frac{3}{5}\)

\(x=\frac{97}{165}\)

vậy \(x=\frac{97}{165}\)

\(x×\left(\frac{3}{7}+\frac{2}{3}\right)=\frac{10}{21}\)

\(x×\frac{23}{21}=\frac{10}{21}\)

\(x=\frac{10}{21}:\frac{23}{21}\)

\(x=\frac{10}{23}\)

vậy \(x=\frac{10}{23}\)

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

=> \(x-\frac{3}{5}=\frac{2}{5}\cdot\left(-\frac{1}{3}\right)=-\frac{2}{15}\)

=> \(x=-\frac{2}{15}+\frac{3}{5}=-\frac{2}{15}+\frac{9}{15}=\frac{7}{15}\)

\(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\)

=> \(\left(\frac{3}{7}-\frac{2}{3}\right)x=\frac{10}{21}\)

=> \(-\frac{5}{21}x=\frac{10}{21}\)

=> \(x=\frac{10}{21}:\frac{-5}{21}=\frac{10}{21}\cdot\frac{-21}{5}=-2\)

Hai bài của ☆luffy cute☆ đều sai hết , xem xét lại đi nhé

1 ) \(\left(x-\frac{3}{5}\right)\div-\frac{1}{3}=\frac{2}{5}\)

\(\Rightarrow x-\frac{3}{5}=\frac{2}{5}\cdot\left(-\frac{1}{3}\right)=-\frac{2}{15}\)

\(\Rightarrow x=-\frac{2}{15}+\frac{3}{5}=-\frac{2}{15}+\frac{9}{15}\)

\(\Rightarrow x=\frac{7}{15}\)

2 ) \(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\)

\(\Rightarrow\left(\frac{3}{7}-\frac{2}{3}\right)x=\frac{10}{21}\)

\(\Rightarrow-\frac{5}{21}x=\frac{10}{21}\)

\(\Rightarrow x=\frac{10}{21}\div-\frac{5}{21}\)

\(\Rightarrow x=-2\)

\(1,\)

\(2x\left(x-3\right)-\left(3-x\right)=0\)

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

\(\Leftrightarrow x^2\left(x^2-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

\(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

a) \(\sqrt{2x-1}=\sqrt{5}\) (ĐK: \(x\ge\dfrac{1}{2}\))

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\left(tm\right)\)

b) \(\sqrt{x-10}=-2\) 

⇒ Giá trị của biểu thức trong căn luôn dương nên phương trình vô nghiệm

c) \(\sqrt{\left(x-5\right)^2}=3\) 

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

TH1: \(\left|x-5\right|=x-5\) với \(x-5\ge0\Leftrightarrow x\ge5\)

Pt trở thành:

\(x-5=3\) (ĐK: \(x\ge5\))

\(\Leftrightarrow x=3+5\)

\(\Leftrightarrow x=8\left(tm\right)\)

TH2: \(\left|x-5\right|=-\left(x-5\right)\) với \(x-5< 0\Leftrightarrow x< 0\)

Pt trở thành:

\(-\left(x-5\right)=3\) (ĐK: \(x< 5\))

\(\Leftrightarrow-x+5=3\)

\(\Leftrightarrow-x=-2\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy: \(S=\left\{2;8\right\}\)

a/ ĐKXĐ: 2x - 1 >= 0 <=> 2x > 1 <=> x>= 1/2

\(\sqrt{2x-1}=\sqrt{5}\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\left(tm\right)\)

b/ ĐKXĐ: x - 10 >= 0 <=> x >= 10

Biểu thức trong căn luôn nhận giá trị dương => vô nghiệm

c/ ĐKXĐ: x - 5 >=0 <=> x >= 5

\(\sqrt{x-5}=3\Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)

Ta có: x²-3x+5 = x²-2.(3/2)x+9/4 + 11/4 = \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\) với mọi x

=> h(x)=x²-3x+5 > 0 với mọi x