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) (x≤3)





 

a: Ta có: \(\sqrt{x-1}=2\)

\(\Leftrightarrow x-1=4\)

hay x=5

b: Ta có: \(\sqrt{3-x}=4\)

\(\Leftrightarrow3-x=16\)

hay x=-13

c: Ta có: \(2\cdot\sqrt{3-2x}=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{3-2x}=\dfrac{1}{4}\)

\(\Leftrightarrow-2x+3=\dfrac{1}{16}\)

\(\Leftrightarrow-2x=-\dfrac{47}{16}\)

hay \(x=\dfrac{47}{32}\)

d: Ta có: \(4-\sqrt{x-1}=\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{2}\)

\(\Leftrightarrow x-1=\dfrac{49}{4}\)

hay \(x=\dfrac{53}{4}\)

e: Ta có: \(\sqrt{x-1}-3=1\)

\(\Leftrightarrow\sqrt{x-1}=4\)

\(\Leftrightarrow x-1=16\)

hay x=17

f:Ta có: \(\dfrac{1}{2}-2\cdot\sqrt{x+2}=\dfrac{1}{4}\)

\(\Leftrightarrow2\cdot\sqrt{x+2}=\dfrac{1}{4}\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{1}{8}\)

\(\Leftrightarrow x+2=\dfrac{1}{64}\)

hay \(x=-\dfrac{127}{64}\)

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

d) \(\left|x-1\right|+\left|x-5\right|+\left|2x+5\right|\)

\(=\left|1-x\right|+\left|5-x\right|+\left|2x+5\right|\)

\(\ge\left|1-x+5-x\right|+\left|2x+5\right|\)

\(\ge\left|6-2x+2x+5\right|=11\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(1-x\right)\left(5-x\right)\ge0\\\left(6-2x\right)\left(2x+5\right)\ge0\end{cases}}\Leftrightarrow-\frac{5}{2}\le x\le1\).

e) \(\left|x+2\right|+\left|x-1\right|+\left|x-4\right|+\left|x+5\right|=12\)

\(\Leftrightarrow\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|=12\)

Có \(\left|x+2\right|+\left|1-x\right|+\left|4-x\right|+\left|x+5\right|\ge\left|x+2+1-x\right|+\left|4-x+x+5\right|=3+9=12\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x+2\right)\left(1-x\right)\ge0\\\left(4-x\right)\left(x+5\right)\ge0\end{cases}}\Leftrightarrow-2\le x\le1\).

f) \(\left|x-1\right|+\left|x-2\right|+\left|x-3\right|+\left|3x-10\right|\)

\(\ge\left|x-1+x-2\right|+\left|3-x+3x-10\right|\)

\(=\left|2x-3\right|+\left|2x-7\right|\)

\(\ge\left|2x-3+7-2x\right|=4\)

Dấu \(=\)khi \(\hept{\begin{cases}\left(x-1\right)\left(x-2\right)\ge0\\\left(3-x\right)\left(3x-10\right)\ge0\\\left(2x-3\right)\left(7-2x\right)\ge0\end{cases}}\Leftrightarrow3\le x\le\frac{10}{3}\).

Lời giải:

$(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$

$\Leftrightarrow 3x^2-2x(a+b+c)+(ab+bc+ac)=0$

Ta thấy:

$\Delta'=(a+b+c)^2-3(ab+bc+ac)=a^2+b^2+c^2-ab-bc-ac$

$=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2}\geq 0$ với mọi $a,b,c\in\mathbb{R}$

$\Rightarrow$ PT đã cho luôn có nghiệm với mọi $a,b,c$

Lời giải:

PT $\Leftrightarrow x^3+9x^2+27x+27-x(9x^2-6x+1)+2x(4x^2-4x+1)=5(1-x^2)$

$\Leftrightarrow x^3+9x^2+27x+27-(9x^3-6x^2+x)+(8x^3-8x+2x)=5-5x^2$

$\Leftrightarrow 12x^2+28x+22=0$

$\Leftrightarrow 6x^2+14x+11=0$

$\Leftrightarrow 6(x+\frac{7}{6})^2=\frac{-17}{6}< 0$ (vô lý)

Vậy PT vô nghiệm.

b)

PT $\Leftrightarrow (x^3+3x^2+3x+1)-(x^3-3x^2+3x-1)-6(x^2-2x+1)=-10$

$\Leftrightarrow 6x^2+2-6x^2+12x-6=-10$

$\Leftrightarrow 12x=-6$

$\Leftrightarrow x=-\frac{1}{2}$

\(3\sqrt{9a^6}-6a^3=3\left|3a^3\right|-6a^3\)

Xét \(a\ge0\Rightarrow\) biểu thức \(=9a^3-6a^3=3a^3\)

Xét \(a< 0\Rightarrow\) biểu thức \(=-9a^3-6a^3=-15a^3\)

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

\(=1-x+3x-1\left(\dfrac{1}{3}< x\le1\right)=2x\)

\(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)=\sqrt{2-\sqrt{3}}.\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)\)

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

\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{2}.\sqrt{3+\sqrt{5}}\)

\(=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{6+2\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2=4^2=16\)

\(\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(2\sqrt{7}-4\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)

\(=2\sqrt{7}-4+\sqrt{7}-1=3\sqrt{7}-5\)

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

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

\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

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

\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

\(=\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}\)

\(=\sqrt{\left(\sqrt{x-4}+2\right)^2}+\sqrt{\left(\sqrt{x-4}-2\right)^2}=\left|\sqrt{x-4}+2\right|+\left|\sqrt{x-4}-2\right|\)

Xét \(x\ge8\Rightarrow\sqrt{x-4}\ge2\Rightarrow\)biểu thức \(=\sqrt{x-4}+2+\sqrt{x-4}-2\)

\(=2\sqrt{x-4}\)

Xét \(x< 8\Rightarrow\sqrt{x-4}< 2\Rightarrow\) biểu thức \(=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)