Bn chụp lại được ko?
Ảnh mờ qé ._.

10a:

\(A=\dfrac{155-\dfrac{10}{7}-\dfrac{5}{11}+\dfrac{5}{23}}{402-\dfrac{26}{7}-\dfrac{13}{11}+\dfrac{13}{23}}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-0,9}{\dfrac{7}{91}+0,2-\dfrac{3}{10}}\)

\(=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{\dfrac{3}{5}+\dfrac{3}{13}-\dfrac{3}{10}}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{1}{10}}\)

\(=\dfrac{5}{13}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{1}{10}\right)}{\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{1}{10}}\)

\(=\dfrac{5}{13}+3=\dfrac{44}{13}\)

Bạn tham khảo nhé.

b)\(\sqrt{x^2-10x+25}=2x-3\)                                               ĐK:x≥3/2

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=2x-3\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=2x-3\\x-5=3-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{8}{3}\end{matrix}\right.\)

Vậy phương trình có 2 nghiệm là ... 

a: Ta có: \(\sqrt{25x+75}+3\sqrt{x-2}=2+4\sqrt{x+3}+\sqrt{9x-18}\)

\(\Leftrightarrow5\sqrt{x+3}-4\sqrt{x+3}=2\)

\(\Leftrightarrow x+3=4\)

hay x=1

c)ĐK:x≥1/2

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

\(\Leftrightarrow\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-\sqrt{2x-1}+1}=2\)

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

\(\Leftrightarrow\left|\sqrt{2x-1}+1\right|+\left|\sqrt{2x-1}-1\right|=2\)

\(\Leftrightarrow\sqrt{2x-1}+1+\left|\sqrt{2x-1}-1\right|=2\)  (do \(\sqrt{2x-1}+1\ge0\))

\(\Leftrightarrow\sqrt{2x-1}+1+\left|1-\sqrt{2x-1}\right|=2\)

Ta có:\(\sqrt{2x-1}+1+\left|1-\sqrt{2x-1}\right|\ge\sqrt{2x-1}+1+1-\sqrt{2x-1}=2\)

Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{2x-1}+1\right)\left(1-\sqrt{2x-1}\right)\ge0\)

                        \(\Leftrightarrow\dfrac{1}{2}\le x\le1\)

Vậy phương trình có tâp nghiệm \(S=\left\{x|\text{​​}\dfrac{1}{2}\le x\le1\right\}\)

26C 27A 28D 29A 30C 31D 32A 33D 34D 35A

mỗi lần đăng chỉ được hỏi 1 bài thôi

bạn đăng nhỏ câu hỏi ra

Bài 2: 

b: \(\left(\dfrac{1}{3}x+3\right)^3=-27\)

\(\Leftrightarrow x\cdot\dfrac{1}{3}+3=-3\)

\(\Leftrightarrow x=-18\)

d: \(2^x+2^{x+3}=144\)

\(\Leftrightarrow2^x\cdot9=144\)

hay x=4

Câu 5:

\(\dfrac{x}{y}=a\Rightarrow\dfrac{x}{a}=\dfrac{y}{1}=\dfrac{x-y}{a-1}=\dfrac{x+y}{a+1}\)

\(\Rightarrow\dfrac{x+y}{x-y}=\dfrac{a+1}{a-1}\)

Câu 6:

\(9x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}\)

\(\Rightarrow\dfrac{x}{5}=\dfrac{y}{9}=\dfrac{3x}{15}=\dfrac{2y}{18}=\dfrac{3x-2y}{15-18}=\dfrac{12}{-3}=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-4\right).5=-20\\y=\left(-4\right).9=-36\end{matrix}\right.\)

Câu 7:

\(\dfrac{x}{-5}=\dfrac{y}{7}=\dfrac{x+y}{-5+7}=\dfrac{-10}{2}=-5\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-5\right).\left(-5\right)=25\\y=\left(-5\right).7=-35\end{matrix}\right.\)