a) P = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-1}\right)\)

P = \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

P = \(\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\dfrac{\left(\sqrt{x}-1\right)+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)

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

P = \(\dfrac{x-1}{\sqrt{x}}\)

b) Để P > 0

\(\Rightarrow\dfrac{x-1}{\sqrt{x}}>0\)

\(x-1>0\)

\(x>1\)

c) Để P = 6

\(\Rightarrow\dfrac{x-1}{\sqrt{x}}=6\)

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

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

\(\left(\sqrt{x}-3-\sqrt{10}\right)\left(\sqrt{x}-3+\sqrt{10}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=3+\sqrt{10}\\\sqrt{x}=3-\sqrt{10}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=19+6\sqrt{10}\\x=19-6\sqrt{10}\end{matrix}\right.\)

1 lít xăng đi dược số km là

54 : 4,5=12(km)

7,5 lít xăng người đó đi đươc số km là

7,5 x 12=90(km)

nhớ tick

Xét \(\Delta ABC\left(\widehat{A}=90^o\right)\) có:

\(\widehat{A}+\widehat{B}+\stackrel\frown{C}=180^o\)

\(90^o+60^o+\widehat{C}=180^o\)

\(\widehat{C}=30^o\)

Ta biết rằng trong một tam giác vuông, cạnh đối diện với góc 30 độ bằng nửa cạnh huyền.

\(\widehat{C}=30^o\) nên cạnh đối diện với góc C là AB = 1/2 cạnh huyền là BC.

Tức là BC = 2AB ( đpcm )

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

\(\Rightarrow a=bk,c=dk\)

Ta có: \(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\left(k-1\right)}=\dfrac{k+1}{k-1}\) (1)

\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\) (2)

Từ (1) và (2) \(\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)

A = \(\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\left(\dfrac{2}{x^2-1}-\dfrac{x}{x-1}+\dfrac{1}{x+1}\right)\)

A = \(\left(\dfrac{x^2+2x+1-x^2+2x-1}{x^2-1}\right):\left(\dfrac{2-x^2-x+x-1}{x^2-1}\right)\)

A = \(\dfrac{4x}{x^2-1}.\dfrac{x^2-1}{1-x^2}\)

A = \(\dfrac{4x}{1-x^2}\)

Gọi số kẹo ban đầu ở hộp A là x, số kẹo ban đầu ở hộp b là 90-x ( x \(\ge\) 0, x < 90 )

Theo bài ra ta có phương trình:

\(x-8=\dfrac{4}{5}\left(90-x+8\right)\)

\(x-8=\dfrac{4}{5}\left(98-x\right)\)

\(x-8=78,4-\dfrac{4}{5}x\)

\(\dfrac{9}{5}x=86,4\)

\(x=48\)( t/m)

Vậy số kẹo ở hộp A là 48 viên, số kẹo ở hộp B là 90 - 48 = 42 viên.

\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)

\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)

\(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

\(\dfrac{18\left(x+7-x-4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)

\(18.3=\left(x+4\right)\left(x+7\right)\)

\(x^2+11x+28-54=0\)

\(x^2+11x-26=0\)

\(\left(x-2\right)\left(x+13\right)=0\)

\(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)

Theo đề x < 0 nên x = -13

P = \(x^4-2x^3+2x^2-2x+1\)

P = \(x^4-x^3-x^3+x^2+x^2-x-x+1\)

P = \(x^3\left(x-1\right)-x^2\left(x-1\right)+x\left(x-1\right)-\left(x-1\right)\)

P = \(\left(x-1\right)\left(x^3-x^2+x-1\right)\)

P = \(\left(x-1\right)\left[x^2\left(x-1\right)+\left(x-1\right)\right]\)

P = \(\left(x-1\right)\left(x-1\right)\left(x^2+1\right)\)

P = \(\left(x-1\right)^2\left(x^2+1\right)\) \(\ge\forall x\) ( đpcm )

Chúc bạn học tốt :))