1, Để hs trên là hàm bậc nhất khi \(m\ne-1\)

Để hàm số trên đồng biến khi \(m+1>0\Leftrightarrow m>-1\)

Để hàm số trên khịch biến khi \(m+1< 0\Leftrightarrow m< -1\)

b, y = (m+1)x - 3 // y = 2x <=> \(\hept{\begin{cases}m+1=2\\-3\ne0\left(luondung\right)\end{cases}\Leftrightarrow m=1}\)

Thay m = 1 vào hs y = (m+1)x - 3 <=> y = 2x - 3 

bạn tự tìm tọa độ rồi tự vẽ nhé vd x = 1 => y = -1 ; x = 2 => y = 1 

Vậy A(1;-1) ; B(2;1) 

\(A=\frac{2}{\sqrt[3]{2}\left(\sqrt[3]{2}^2+\sqrt[3]{2}+1\right)}=\frac{2\left(\sqrt[3]{2}-1\right)}{\sqrt[3]{2}\left(\sqrt[3]{2}^2+\sqrt[3]{2}+1\right)\left(\sqrt[3]{2}-1\right)}=\frac{2\left(\sqrt[3]{2}-1\right)}{\sqrt[3]{2}}=2-\sqrt[3]{4}\)

hôm qua mình làm B rồi nhé 

\(P=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+\sqrt{x}}\)ĐK : x > 0 

\(=\frac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}:\frac{\sqrt{x}}{x+\sqrt{x}}=\sqrt{x}+1+\frac{1}{\sqrt{x}}\)

\(P=\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{3}{\sqrt{x}+1}-\frac{6\sqrt{x}-4}{x-1}\)Với x >= 0 ; \(x\ne1\)

\(=\frac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{x-1}=\frac{x-2\sqrt{x}+1}{x-1}=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)

CẢM MƠN ANH TÚ NHIỀU Ạ

\(\hept{\begin{cases}x\left(x+y\right)-4=-x\\\left(x^2+2xy+y^2\right)-\frac{5}{x^2}=4\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+y\right)-\frac{4}{x}=-1\\\left(x^2+2xy+y^2\right)-\frac{5}{x^2}=4\end{cases}}}\)

Đặt \(\hept{\begin{cases}x+y=a\\\frac{1}{x}=b\end{cases}}\Rightarrow\hept{\begin{cases}a-4b=-1\\a^2-5b^2=4\end{cases}\Rightarrow a=4b-1}\)

\(\Rightarrow a^2-5b^2=4\Leftrightarrow\left(4a-1\right)^2-5b^2=4\Leftrightarrow\orbr{\begin{cases}b=1\Rightarrow a=3\\b=-\frac{3}{11}\Rightarrow a=-\frac{23}{11}\end{cases}}\)

vậy \(\orbr{\begin{cases}x=1,y=2\\x=-\frac{11}{3},y=\frac{5}{3}\end{cases}}\)

\(\sqrt[3]{3a}-6\sqrt[3]{\frac{a}{9}}+a\sqrt[3]{\frac{3}{a^2}}-\sqrt[3]{5\sqrt{5}}=\sqrt[3]{3a}-2\sqrt[3]{\frac{3^3}{9}a}+\sqrt[3]{\frac{3a^3}{a^2}}-\sqrt[3]{\sqrt{5}^3}\)

\(=\sqrt[3]{3a}-2\sqrt[3]{3a}+\sqrt[3]{3a}+\sqrt{5}=\sqrt{5}\)

\(\sqrt{x+\frac{1}{4}}+\frac{1}{2}=\sqrt{2}\Leftrightarrow\sqrt{x+\frac{1}{4}}=\sqrt{2}-\frac{1}{2}\)ĐK : x >= -1/4 

\(\Leftrightarrow x+\frac{1}{4}=2-\sqrt{2}+\frac{1}{4}\Leftrightarrow x=2-\sqrt{2}\)(tm)

tìm x giúp mình với cảm ơn

\(4\sqrt[3]{x^2-6x+10}-3=1\Leftrightarrow4\sqrt[3]{x^2-6x+10}=4\)

\(\Leftrightarrow\sqrt[3]{x^2-6x+10}=1\Leftrightarrow x^2-6x+10=1\)

\(\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x=3\)

\(4\sqrt[3]{x^2-6x+10}-3=1\)

\(\Leftrightarrow4\sqrt[3]{x^2-6x+10}=4\)

\(\Leftrightarrow\sqrt[3]{x^2-6x+10}=1\)

\(\Leftrightarrow x^2-6x+10=1\)

\(\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x=3\)

a,đk x >= 0  \(\sqrt{16x}=8\Leftrightarrow4\sqrt{x}=8\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

b, đk x =< 4/5 \(\sqrt{4-5x}=12\Leftrightarrow4-5x=144\Leftrightarrow5x=-140\Leftrightarrow x=-28\)

c;d;e tương tự câu f bạn nhé 

f, đk x >= -1 

\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}=16-\sqrt{x+1}\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\Leftrightarrow\sqrt{x+1}=4\Leftrightarrow x=15\)

Câu 33 : \(\sqrt[3]{x^3+3x^2+3x+1}-\sqrt[3]{8x^3+12x^2+6x+1}\)

\(=\sqrt[3]{\left(x+1\right)^3}-\sqrt[3]{\left(2x+1\right)^3}=x+1-2x-1=-x\)

-> chọn B 

Câu 34 : \(\sqrt[3]{x^3-3x^2+3x-1}-\sqrt[3]{125x^3+75x^2+15x+1}\)

\(=\sqrt[3]{\left(x-1\right)^3}-\sqrt[3]{\left(5x+1\right)^3}=x-1-5x-1=-4x-2\)

ta có : \(\hept{\begin{cases}x^3+3x^2+3x+1=\left(x+1\right)^3\\8x^3+12x^2+6x+1=\left(2x+1\right)^3\end{cases}}\)

nên : \(\sqrt[3]{x^3+3x^2+3x+1}-\sqrt[3]{8x^3+12x^2+6x+1}=x+1-\left(2x+1\right)=-x\)

Vậy đáp án là B