?

câu nào

a: Xét (O) có 

ΔABC nội tiếp đường tròn

AB là đường kính

Do đó: ΔABC vuông tại C

Câu 1:

1) Ta có: \(2x^2+5x-3=0\)

\(\Leftrightarrow2x^2+6x-x-3=0\)

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

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

2) Để hàm số đồng biến trên R thì m-1>0

hay m>1

Câu 1: 

3) Ta có: P=a+b-2ab

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

\(=2-2\cdot\left(-1\right)=4\)

Giúp em vs m.n

c: \(=5\cdot\dfrac{1}{5}-3\cdot\dfrac{1}{3}=0\)

\(\text{b)}\left(\dfrac{3}{5}\right)^2-\left[\dfrac{1}{3}:3-\sqrt{16}.\left(\dfrac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)

\(=\dfrac{9}{25}-\left[\dfrac{1}{9}-1\right]-1\)

\(=\dfrac{9}{25}-\left(\dfrac{-8}{9}\right)-1\)

\(=\dfrac{281}{225}-1\)

\(=\dfrac{56}{225}\)

\(\text{c)}5\sqrt{\dfrac{1}{25}}-3\sqrt{\dfrac{1}{9}}\)

\(=5.\dfrac{1}{5}-3.\dfrac{1}{3}\)

\(=1-1\)

\(=0\)

2:

a: T=900-30c

b: giá trị của 900 thùng là:

900*2000000=1800000000(đồng)

Số tiền phải chi là:

900:30*2500000=75000000(đồng)

Xưởng sẽ lời khoảng:

1800000000-75000000=1725000000(đồng)

phương thức biểu đạt : biểu cảm " chưa chắc lắm "

thể thơ 5 chữ

1.

b, \(B=\dfrac{8+2\sqrt{2}}{3-\sqrt{2}}-\dfrac{2+3\sqrt{2}}{\sqrt{2}}+\dfrac{\sqrt{2}}{1-\sqrt{2}}\)

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

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

\(=-1\)

Bài 1: 

b: Ta có: \(B=\dfrac{8+2\sqrt{2}}{3-\sqrt{2}}-\dfrac{2+3\sqrt{2}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{2}-1}\)

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

\(=4+2\sqrt{2}-5\)

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

4.

a, ĐK: \(x\ge0;x\ne1\)

\(Q=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right).\dfrac{\left(1-x\right)^2}{2}\)

\(=\left[\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}\)

\(=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}\)

\(=\left[\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\dfrac{\left(1-x\right)^2}{2}\)

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

\(=\dfrac{\sqrt{x}\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(=\sqrt{x}-x\)

b, \(Q=\sqrt{x}-x=\sqrt{x}\left(1-\sqrt{x}\right)>0\)

\(\Leftrightarrow1-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 1\)

\(\Leftrightarrow x< 1\)

Vậy \(0\le x< 1\)

c, \(Q=\sqrt{x}-x\)

\(=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}\)

\(=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)

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