a) \(6\left(1,5-2x\right)=3\left(-15+2x\right)\)

\(\Rightarrow6.1,5-6.2x=3.\left(-15\right)+3.2x\)

\(\Rightarrow9-12x=-45+6x\)

\(\Rightarrow9-12x+45-6x=0\)

\(\Rightarrow54-18x=0\)

\(\Rightarrow18\left(3-x\right)=0\)

Để 18(3 - x) = 0

=> 3 - x = 0

=> x = 3

Vậy nghiệm của phương trình là 3

b) \(3-4x\left(25-2x\right)=8x^2+x-300\)

\(\Rightarrow3-100x+8x^2=8x^2+x-300\)

\(\Rightarrow3-100x+8x^2-8x^2-x+300=0\)

\(\Rightarrow303-101x=0\)

\(\Rightarrow101\left(3-x\right)=0\)

Để 101(3 - x) = 0

=> 3 - x = 0

=> x = 3

Vậy nghiệm của phương trình là 3

c) \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{16}{x^2-1}\)

\(\Rightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{16}{x^2-1}\)

\(\Rightarrow\dfrac{\left(x+1\right)^2}{x^2-1}-\dfrac{\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)

\(\Rightarrow\dfrac{\left(x+1\right)^2-\left(x-1\right)^2}{x^2-1}=\dfrac{16}{x^2-1}\)

\(\Rightarrow\dfrac{\left(x+1+x-1\right)\left(x+1-x+1\right)}{x^2-1}=\dfrac{16}{x^2-1}\)

\(\Rightarrow\dfrac{2x.2}{x^2-1}-\dfrac{16}{x^2-1}=0\)

\(\Rightarrow\dfrac{4x-16}{x^2-1}=0\)

\(\Rightarrow4x-16=0\)

\(\Rightarrow4\left(x-4\right)=0\)

Để 4(x - 4) = 0

=> x - 4 = 0

=> x = 4

Vậy nghiệm của phương trình là 4

d) \(x^2-x-6=0\)

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

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

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

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

Vậy nghiệm của phương trình là -2;3

@Mysterious Person @Aki Tsuki @Nhã Doanh @Phùng Khánh Linh giúp vs! cần gấp lắm!

\(2\cdot\frac{x}{3}-\frac{1}{2}\cdot x=1.5\)

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

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

\(x\cdot\left(\frac{2}{3}-\frac{1}{2}\right)=5\)

\(x\cdot\frac{1}{6}=5\)

\(x=5:\frac{1}{6}\)

\(x=5\cdot6\)

\(x=30\)

ai k mh mh k lại

k cho mh nha bạn

\(\left|x+\dfrac{1}{1.5}\right|+\left|x+\dfrac{1}{5.9}\right|+\left|x+\dfrac{1}{9.14}\right|+...+\left|x+\dfrac{1}{397.401}\right|\ge0\)

\(\Rightarrow101x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow x+\dfrac{1}{1.5}+x+\dfrac{1}{5.9}+...+x+\dfrac{1}{397.401}=101x\)

\(\Rightarrow101x+\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{397.401}\right)=x\)

\(\Rightarrow\dfrac{1}{4}\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{397.401}\right)=x\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+....+\dfrac{1}{397}-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}\left(1-\dfrac{1}{401}\right)\)

\(\Rightarrow x=\dfrac{1}{4}.\dfrac{400}{401}\)

\(\Rightarrow x=\dfrac{100}{401}\)

a.

- Với \(m=\pm1\Rightarrow-6x=1\Rightarrow x=-\dfrac{1}{6}\) có nghiệm

Đặt \(f\left(x\right)=\left(1-m^2\right)x^3-6x-1\)

- Với \(\left[{}\begin{matrix}m>1\\m< -1\end{matrix}\right.\Rightarrow1-m^2>0\)

\(f\left(0\right)=-1< 0\)

\(\lim\limits_{x\rightarrow-\infty}f\left(x\right)=\lim\limits_{x\rightarrow-\infty}\left[\left(1-m\right)^2x^3-6x-1\right]\)

\(=\lim\limits_{x\rightarrow-\infty}x^3\left(1-m^2-\dfrac{6}{m^2}-\dfrac{1}{m^3}\right)=-\infty\left(1-m^2\right)=+\infty\) dương

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(-\infty;0\right)\)

- Với \(-1< m< 1\Rightarrow1-m^2< 0\)

\(\lim\limits_{x\rightarrow+\infty}\left[\left(1-m^2\right)x^3-6x-1\right]=\lim\limits_{x\rightarrow+\infty}x^3\left[\left(1-m^2\right)-\dfrac{6}{x^2}-\dfrac{1}{x^3}\right]=+\infty\left(1-m^2\right)=+\infty\) dương

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc \(\left(0;+\infty\right)\)

Vậy pt đã cho có nghiệm với mọi m

b. Để chứng minh pt này có đúng 1 nghiệm thì cần áp dụng thêm kiến thức 12 (tính đơn điệu của hàm số). Chỉ bằng kiến thức 11 sẽ ko chứng minh được

c. 

Đặt \(f\left(x\right)=\left(m-1\right)\left(x-2\right)^2\left(x-3\right)^3+2x-5\)

Do \(f\left(x\right)\) là hàm đa thức nên \(f\left(x\right)\) liên tục trên R

\(f\left(2\right)=4-5=-1< 0\)

\(f\left(3\right)=6-5=1>0\)

\(\Rightarrow f\left(2\right).f\left(3\right)< 0\) với mọi m

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm thuộc (2;3) với mọi m

Hay pt đã cho luôn luôn có nghiệm

a, \(3\dfrac{1}{2}-\dfrac{2}{3}:\left(1-2x\right)=1\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{2}{3}:\left(1-2x\right)=\dfrac{7}{2}-\dfrac{3}{2}=2\)

\(\Leftrightarrow1-2x=\dfrac{2}{3}:2=\dfrac{1}{3}\Leftrightarrow2x=1-\dfrac{1}{3}=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{3}\)

b, pt \(\Leftrightarrow\dfrac{-1,5}{3\left|2x+1\right|}=\dfrac{-1}{14}\Rightarrow3\left|2x+1\right|=21\Rightarrow\left|2x+1\right|=7\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=-7\\2x-1=7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)

\(a,3\dfrac{1}{2}-\dfrac{2}{3}:\left(1-2x\right)=1\dfrac{1}{2}\)

\(\Rightarrow\dfrac{2}{3}:\left(1-2x\right)=3\dfrac{1}{2}-1\dfrac{1}{2}\)

\(\Rightarrow\dfrac{2}{3}:\left(1-2x\right)=2\)

\(\Rightarrow1-2x=\dfrac{2}{3}:2\)

\(\Rightarrow1-2x=\dfrac{1}{3}\)

\(\Rightarrow2x=1-\dfrac{1}{3}\)

\(\Rightarrow2x=\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{\dfrac{2}{3}}{2}\Rightarrow x=\dfrac{2}{3}.\dfrac{1}{2}\Rightarrow x=\dfrac{1}{3}\)

Vậy \(x=\dfrac{1}{3}\)

\(b,\dfrac{\left(-1\right).5}{3\left|2x+1\right|}=\dfrac{\sqrt{64}}{-112}\)

Từ đẳng thức:

\(\Rightarrow3\left|2x+1\right|.\sqrt{64}=\left[\left(-1\right).5\right].\left(-112\right)\)

\(\Rightarrow3\left|2x+1\right|.8=\left[\left(-1\right).5\right].\left(-112\right)\)

\(\Rightarrow3\left|2x+1\right|.8=\left(-5\right).\left(-112\right)\)

\(\Rightarrow3\left|2x+1\right|.8=560\)

\(\Rightarrow\left|2x+1\right|=\dfrac{560}{\dfrac{8}{7}}\)

\(\Rightarrow\left|2x+1\right|=10\)

Ta có:

\(\left\{{}\begin{matrix}2x+1=10\\2x+1=\left(-10\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{10-1}{2}\\x=\dfrac{\left(-10\right)-1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=4,5\\x=-5.5\end{matrix}\right.\)

Vậy \(x\in\left\{-5,5;4,5\right\}\)

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a/ \(A=\frac{x}{2}+\frac{1}{2x}+\frac{5x}{2}\ge2\sqrt{\frac{x}{4x}}+\frac{5}{2}.1=\frac{7}{2}\)

\("="\Leftrightarrow x=1\)

b/ \(B=\frac{3\left(x+1\right)}{2}+\frac{1}{x+1}-\frac{3}{2}\ge2\sqrt{\frac{3\left(x+1\right)}{2\left(x+1\right)}}-\frac{3}{2}=\frac{-3+2\sqrt{6}}{2}\)

\("="\Leftrightarrow\left(x+1\right)^2=\frac{2}{3}\Rightarrow x=...\)

c/ \(C=\frac{2x-1}{6}+\frac{5}{2x-1}+\frac{1}{6}\ge2\sqrt{\frac{5\left(2x-1\right)}{6\left(2x-1\right)}}+\frac{1}{6}=\frac{1+2\sqrt{30}}{6}\)

\("="\Leftrightarrow\left(2x-1\right)^2=30\Rightarrow x=...\)

d/ \(D=x+\frac{4}{x}+4\ge2\sqrt{\frac{4x}{x}}+4=8\)

\("="\Leftrightarrow x^2=4\Rightarrow x=...\)

e/ \(E=\left(x+3\right)\left(5-x\right)\le\frac{1}{4}\left(x+3+5-x\right)^2=16\)

\("="\Leftrightarrow x+3=5-x\Rightarrow x=...\)

f/ \(F=\frac{1}{2}\left(2x+6\right)\left(5-2x\right)\le\frac{1}{8}\left(2x+6+5-2x\right)^2=\frac{121}{8}\)

\("="\Leftrightarrow2x+6=5-2x\Leftrightarrow x=...\)

\(M=-\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{n\left(n+4\right)}\right)\\ =-\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n}-\dfrac{1}{n+4}\right)\\ =-\left(1-\dfrac{1}{n+4}\right)\\ =-\left(\dfrac{n+3}{n+4}\right)\)

\(-4x\left(x-5\right)-2x\left(8-2x\right)=-3\\ \Rightarrow-4x^2+20x-16x+4x^2=-3\\ \Rightarrow4x=-3\\ \Rightarrow x=-\dfrac{3}{4}\)

a) \(M=-\dfrac{4}{1.5}-\dfrac{4}{5.9}-...-\dfrac{4}{\left(n-4\right)n}\)

⇔ \(M=-\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{\left(n-4\right)n}\right)\)

⇔ \(M=-\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{n-4}-\dfrac{1}{n}\right)\)

⇔ \(M=-\left(1-\dfrac{1}{n}\right)\)

⇔ \(M=-\dfrac{n-1}{n}\)

b) - 4x(x - 5) - 2x(8 - 2x) = -3

⇔ -4x² + 20x - 16x + 4x² = -3

⇔ 4x = -3

⇔ x = \(-\dfrac{3}{4}\)