d: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\left(3-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)

\(=\dfrac{5}{6}\cdot\dfrac{6}{5}-17=1-17=-16\)

h: \(\dfrac{\left(-1\right)^3}{15}+\left(-\dfrac{2}{3}\right)^2:2\dfrac{2}{3}-\left|-\dfrac{5}{6}\right|\)

\(=-\dfrac{1}{15}+\dfrac{-8}{27}:\dfrac{8}{3}-\dfrac{5}{6}\)

\(=-\dfrac{1}{15}-\dfrac{1}{9}-\dfrac{5}{6}\)

\(=\dfrac{-6-10-75}{90}=\dfrac{-91}{90}\)

k: \(\dfrac{2\cdot6^9-2^5\cdot18^4}{2^2\cdot6^8}\)
\(=\dfrac{2^{10}\cdot3^9-2^5\cdot2^4\cdot3^8}{2^2\cdot2^8\cdot3^8}\)

\(=\dfrac{2^{10}\cdot3^9-2^9\cdot3^8}{2^{10}\cdot3^8}=\dfrac{2^9\cdot3^8\left(2\cdot3-1\right)}{2^{10}\cdot3^8}\)

\(=\dfrac{5}{2}\)

n: \(3-\left(-\dfrac{7}{8}\right)^0+\left(\dfrac{1}{2}\right)^3\cdot16\)

\(=3-1+\dfrac{1}{8}\cdot16\)

=2+2

=4

1.

Áp dụng BĐT Cauchy-Schwarz:

\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)

Tương tự:

\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)

\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)

Cộng vế:

\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)

Dấu "=" xảy ra khi \(a=b=c\)

2.

Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)

Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)

Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)

Biến đổi giả thiết:

\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)

\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)

\(\Rightarrow ab+bc+ca=a+b+c-1\)

BĐT cần chứng minh trở thành:

\(a^2+b^2+c^2\ge1\)

\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)

\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)

\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)

Giúp mk vs ,mk cần gấp

a) Ta có: \(A=\left(\dfrac{2}{x+2}-\dfrac{1}{x-3}+\dfrac{5-x}{x^2-x-6}\right)\cdot\left(x-\dfrac{6}{x-1}\right)\)

\(=\left(\dfrac{2\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}-\dfrac{x+2}{\left(x-3\right)\left(x+2\right)}+\dfrac{5-x}{\left(x-3\right)\left(x+2\right)}\right)\cdot\dfrac{x\left(x-1\right)-6}{x-1}\)

\(=\dfrac{2x-6-x-2+5-x}{\left(x+2\right)\left(x-3\right)}\cdot\dfrac{x^2-x-6}{x-1}\)

\(=\dfrac{-3}{x-1}\)

Bài 1:

a.

$|x+\frac{7}{4}|=\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)

b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$

$|2x+1|=\frac{11}{15}$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)

c.

$3x(x+\frac{2}{3})=0$

\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)

d.

$x+\frac{1}{3}=\frac{2}{5}-(\frac{-1}{3})=\frac{2}{5}+\frac{1}{3}$

$\Leftrightarrow x=\frac{2}{5}$

Bài 2:

$\frac{1}{100}-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}$

$=\frac{99}{100}$

$\Rightarrow A=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=\frac{-49}{50}$

Bài 1: 

a) Ta có: \(\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{1}{2}\\x+\dfrac{7}{4}=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)

b) Ta có: \(\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}\)

\(\Leftrightarrow\left|2x+1\right|=\dfrac{11}{15}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{11}{15}\\2x+1=\dfrac{-11}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-4}{15}\\2x=\dfrac{-26}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{15}\\x=\dfrac{-13}{15}\end{matrix}\right.\)

c) Ta có: \(3x\left(x+\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)

(a) Với \(x\ge0,x\ne9\), ta có: \(A=\left(\dfrac{2}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{2\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{3}{\sqrt{x}+3}.\)

(b) Ta có: \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)

\(\Rightarrow\sqrt{x}=2+\sqrt{3}\).

Thay vào biểu thức \(A\) (thỏa mãn điều kiện), ta được: \(A=\dfrac{3}{2+\sqrt{3}+3}=\dfrac{3}{5+\sqrt{3}}\)

\(=\dfrac{3\left(5-\sqrt{3}\right)}{5^2-\left(\sqrt{3}\right)^2}=\dfrac{15-3\sqrt{3}}{22}.\)

(c) Để \(A=\dfrac{3}{5}\Rightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{3}{5}\)

\(\Rightarrow\sqrt{x}+2=5\Leftrightarrow x=9\) (không thỏa mãn).

Vậy: \(x\in\varnothing.\)

(d) Để \(A>1\Leftrightarrow A-1>0\Rightarrow\dfrac{3}{\sqrt{x}+3}-1>0\)

\(\Leftrightarrow\dfrac{1-\sqrt{x}}{\sqrt{x}+3}>0\Rightarrow1-\sqrt{x}>0\) (do \(\sqrt{x}+3>0\forall x\inĐKXĐ\))

\(\Rightarrow x< 1\). Kết hợp với điều kiện thì \(0\le x< 1.\)

(e) \(A\in Z\Rightarrow\dfrac{3}{\sqrt{x}+3}\in Z\Rightarrow\left(\sqrt{x}+3\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+3=1\\\sqrt{x}+3=-1\\\sqrt{x}+3=3\\\sqrt{x}+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-2\left(VL\right)\\\sqrt{x}=-4\left(VL\right)\\\sqrt{x}=0\Leftrightarrow x=0\left(TM\right)\\\sqrt{x}=-6\left(VL\right)\end{matrix}\right.\)

Vậy: \(x=0.\)

Bài 1:

\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)

Bài  2:

\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)

Bổ sung phần c và d luôn:

c, C = \(\dfrac{2}{5}\)

\(\Leftrightarrow\) \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{2}{5}\)

\(\Leftrightarrow\) 5(x² - 1) = 2(2x² + 3)

\(\Leftrightarrow\) 5x² - 5 = 4x² + 6

\(\Leftrightarrow\) x² = 11

\(\Leftrightarrow\) x² - 11 = 0

\(\Leftrightarrow\) (x - \(\sqrt{11}\))(x + \(\sqrt{11}\)) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-\sqrt{11}=0\\x+\sqrt{11}=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\sqrt{11}\left(TM\right)\\x=-\sqrt{11}\left(TM\right)\end{matrix}\right.\)

d, Ta có: \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{x^2+\dfrac{3}{2}-\dfrac{5}{2}}{2\left(x^2+\dfrac{3}{2}\right)}\) = \(\dfrac{1}{2}\) - \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\)

C nguyên \(\Leftrightarrow\) \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\) nguyên \(\Leftrightarrow\) 5 \(⋮\) 4(x² + \(\dfrac{3}{2}\))

\(\Leftrightarrow\) 4(x² + \(\dfrac{3}{2}\)) \(\in\) Ư(5)

Xét các TH:

4(x² + \(\dfrac{3}{2}\)) = 5 \(\Leftrightarrow\) x² = \(\dfrac{-1}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{1}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = -5 \(\Leftrightarrow\) x² = \(\dfrac{-11}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{11}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = 1 \(\Leftrightarrow\) x² = \(\dfrac{-5}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{5}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = -1 \(\Leftrightarrow\) x² = \(\dfrac{-7}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{7}{4}\) = 0 (Vô nghiệm)

Vậy không có giá trị nào của x \(\in\) Z thỏa mãn C \(\in\) Z

Chúc bn học tốt! (Ko bt đề sai hay ko nữa :v)