Giả sử Amy trả lời đúng tất cả các câu hỏi, số điểm mà Amy sẽ đạt được là: \(5\times30=150\) (điểm)

Số điểm nhiều hơn với thực tế là: \(150-122=28\) (điểm)

Số điểm trả lời đúng nhiều hơn số điểm trả lời sai là: \(5+2=7\) (điểm)

Số câu trả lời sai: \(28\div7=4\) (câu)

Số câu trả lời đúng: \(30-4=26\) (câu)

Amy trả lời được 26 câu bạn nhé

2⁷ : 2² = 2⁵ = 32

5⁴ : 5³ = 5 

2⁴ - 3.2⁵ = 16 - 96 = - 80 

1) \(\sqrt{4+x}=2-x\) (ĐK: \(x\ge-4\))

\(\Leftrightarrow4+x=\left(2-x\right)^2\)

\(\Leftrightarrow4+x=4-4x+x^2\)

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

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

Vậy: \(S=\left\{0;5\right\}\)

2) 

a) ĐKXĐ: \(a>0,a\ne1\)

\(A=\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{a+\sqrt{a}}\right):\dfrac{\sqrt{a}+1}{a}\)

\(A=\left[\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}-\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}+1\right)}\right]\cdot\dfrac{a}{\sqrt{a}+1}\)

\(A=\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\cdot\dfrac{a}{\sqrt{a}+1}\)

\(A=\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a}{\sqrt{a}+1}\)

\(A=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}}\cdot\dfrac{\sqrt{a}\cdot\sqrt{a}}{\sqrt{a}+1}\)

\(A=\sqrt{a}\left(\sqrt{a}-1\right)\)

\(A=a-\sqrt{a}\)

b) Ta có:

\(A=a-\sqrt{a}\)

\(A=\left(\sqrt{a}\right)^2-2\cdot\dfrac{1}{2}\cdot\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}\)

\(A=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)

Mà: \(\left(\sqrt{a}-\dfrac{1}{2}\right)^2\ge0\) nên \(A=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)

Dấu "=" xảy ra khi:

\(\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}=-\dfrac{1}{4}\)

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

Vậy: \(A_{min}=-\dfrac{1}{4}\)khi \(a=\dfrac{1}{4}\)

Ta có:

VT: \(\left(xy+1\right)\left(x^2y^2-xy+1\right)+\left(x^3-1\right)\left(1-y^3\right)\)

\(=\left(xy\right)^3+1^3+x^3-x^3y^3-1+y^3\)

\(=x^3y^3+1+x^3-x^3y^3-1+y^3\)

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

\(=x^3+y^3=VP\left(dpcm\right)\)

a) \(\sqrt{x}-x-0\) (ĐK: \(x\ge0\))

\(\Leftrightarrow\sqrt{x}\left(1-\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\1-\sqrt{x}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

b) \(x-\sqrt{2x-9}=6\)

\(\Leftrightarrow\sqrt{2x-9}=x-6\) (ĐK: \(x\ge\dfrac{9}{2}\))

\(\Leftrightarrow2x-9=\left(x-6\right)^2\)

\(\Leftrightarrow2x-9=x^2-12x+36\)

\(\Leftrightarrow x^2-14x+45=0\)

\(\Leftrightarrow x^2-5x-9x+45=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\)

c) \(3x-\sqrt{6x-\left(3-2\right)}=0\) (ĐK: \(x\ge\dfrac{1}{6}\))

\(\Leftrightarrow3x-\sqrt{6x-1}=0\)

\(\Leftrightarrow\sqrt{6x-1}=3x\)

\(\Leftrightarrow6x-1=9x^2\)

\(\Leftrightarrow9x^2-6x+1=0\)

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

\(\Leftrightarrow x=\dfrac{1}{3}\left(tm\right)\)

\(a^2+16\cdot a\)

\(=a\cdot\left(a-16\right)\)

Bạn xem lại đề

a) \(\left(\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5\right):\dfrac{3}{5}ax^3\)

\(=\dfrac{\dfrac{3}{5}a^6x^3+\dfrac{3}{7}a^3x^4-\dfrac{9}{10}ax^5}{\dfrac{3}{5}ax^3}\)

\(=\dfrac{\dfrac{3}{5}ax^3\left(a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}x^2\right)}{\dfrac{3}{5}ax^3}\)

\(=a^5+\dfrac{5}{7}a^2x-\dfrac{3}{2}x^2\)

b) \(\left(9x^2y^3-15x^4y^4\right):3x^2y-\left(2-3x^2y\right)\cdot y^2\)

\(=\dfrac{3x^2y\left(3y^2-5x^2y^3\right)}{3x^2y}-2y^2+3x^2y^3\)

\(=3y^2-5x^2y^3-2y^2+3x^2y^3\)

\(=y^2-2x^2y^3\)

c) \(\left(6x^2-xy\right):x+\left(2x^3y+3xy^2\right):xy-\left(2x-1\right)x\)

\(=\dfrac{6x^2-xy}{x}+\dfrac{2x^3y+3xy^2}{xy}-x\left(2x-1\right)\)

\(=\dfrac{x\left(6x-y\right)}{x}+\dfrac{xy\left(2x^2+3y\right)}{xy}-2x^2+x\)

\(=6x-y+2x^2+3y-2x^2+x\)

\(=7x+2y\)

d) \(\left(x^2-xy\right):x+\left(6x^2y^5-9x^3y^4+15x^4y^2\right):\dfrac{3}{2}x^2y^3\)

\(=\dfrac{x^2-xy}{x}+\dfrac{6x^2y^5-9x^3y^4+15x^4y^2}{\dfrac{3}{2}x^2y^3}\)

\(=\dfrac{x\left(x-y\right)}{x}+\dfrac{\dfrac{3}{2}x^2y^2\left(4y^3-6xy^2+10x^2\right)}{\dfrac{3}{2}x^2y^3}\)

\(=x-y+\dfrac{4y^3-6xy^2+10x^2}{y}\)

\(D=1-4+4^2-4^3+4^4-...+4^{2016}-4^{2017}\) (sửa \(2^{2017}\) thành \(4^{2017}\))

\(\Rightarrow D=\left(1-4\right)+4^2\left(1-4\right)+4^4\left(1-4\right)-...+4^{2016}\left(1-4\right)\)

\(\Rightarrow D=\left(-3\right)+4^2.\left(-3\right)+4^4.\left(-3\right)-...+4^{2016}.\left(-3\right)\)

\(\Rightarrow D=\left(-3\right)\left(1+4^2+4^4+...+4^{2016}\right)\)

\(\Rightarrow4D=\left(-3\right)\left(4+4^3+4^5+...+4^{2017}\right)\)

\(\Rightarrow4D+D=\left(-3\right)\left(1+4+4^2+4^3+4^4+...+4^{2016}+4^{2017}\right)\)

\(\Rightarrow5D=\left(-3\right)\dfrac{4^{2017+1}-1}{4-1}\)

\(\Rightarrow D=\left(-3\right)\dfrac{4^{2018}-1}{3.5}\)

\(\Rightarrow D=\left(-1\right)\dfrac{4^{2018}-1}{5}=\dfrac{1-4^{2018}}{5}\)

\(A=2+2^2+...2^{2021}\)

\(\Rightarrow A+1=1+2+2^2+...2^{2021}\)

\(\Rightarrow A+1=\dfrac{2^{2021+1}-1}{2-1}\)

\(\Rightarrow A+1=2^{2022}-1\)

\(\Rightarrow A=2^{2022}-2< 2^{2022}=B\)

\(\Rightarrow A< B\)

Ta có:

P = a + b + c ≤ a + b + a + b = 2(a + b) ≤ 2(-1) = -2

Ta cũng có:

P = a + b + c ≤ a + b + c - 2abc ≥ a + b + c - 2(-1)(-1)(-1) = -3

Vậy GTNN của P = -3 và GTLN của P = -2.