Bạn ơi chung căn (3x-2) hãy căn (3x) - 2

Lời giải:

\(-3x^2+8x-2=0\)

\(\Leftrightarrow 3x^2-8x+2=0\)

\(\Leftrightarrow 3(x^2-\frac{8}{3}x+\frac{8^2}{6^2})=\frac{10}{3}\)

\(\Leftrightarrow 3(x-\frac{8}{6})^2=\frac{10}{3}\)

\(\Leftrightarrow (x-\frac{4}{3})^2=\frac{10}{9}\Rightarrow \left[\begin{matrix} x-\frac{4}{3}=\frac{\sqrt{10}}{3}\\ x-\frac{4}{3}=\frac{-\sqrt{10}}{3}\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=\frac{4+\sqrt{10}}{3}\\ x=\frac{4-\sqrt{10}}{3}\end{matrix}\right.\)

\(-3x^2+8x-2=0\)

\(\Leftrightarrow-3\left(x^2-\dfrac{8}{3}x+\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow-3\left(x^2-2x.\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{10}{9}\right)=0\)

\(\Leftrightarrow-3\left[\left(x-\dfrac{4}{3}\right)^2-\dfrac{10}{9}\right]=0\)

\(\Leftrightarrow\left(x-\dfrac{4}{3}\right)^2-\dfrac{10}{9}=0\)

\(\Leftrightarrow\left(x-\dfrac{4}{3}\right)^2=\dfrac{10}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{4}{3}=\dfrac{\sqrt{10}}{3}\\x-\dfrac{4}{3}=\dfrac{-\sqrt{10}}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{10}+4}{3}\\x=\dfrac{4-\sqrt{10}}{3}\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=\dfrac{\sqrt{10}+4}{3}\\x=\dfrac{4-\sqrt{10}}{3}\end{matrix}\right.\)

TH1:  \(m=-1\) thỏa mãn (dễ dàng kiểm tra các giá trị \(f\left(-1\right)>0\) ; \(f\left(0\right)< 0\) ; \(f\left(3\right)>0\) nên pt có ít nhất 2 nghiệm thuộc (-1;0) và (0;3)

TH2: \(m>-1\):

\(\lim\limits_{x\rightarrow+\infty}f\left(x\right)=\lim\limits_{x\rightarrow+\infty}x^4\left[m\left(1-\dfrac{2}{x}\right)^2\left(1+\dfrac{9}{x}\right)+1-\dfrac{32}{x^4}\right]=+\infty.\left(m+1\right)=+\infty>0\)

\(\Rightarrow\) Luôn tồn tại 1 giá trị \(x=a\) đủ lớn sao cho \(f\left(a\right)>0\)

\(f\left(0\right)=-32< 0\Rightarrow f\left(a\right).f\left(0\right)< 0\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm dương

\(f\left(-9\right)=9^4-32>0\Rightarrow f\left(-9\right).f\left(0\right)< 0\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 1 nghiệm âm thuộc \(\left(-9;0\right)\)

\(\Rightarrow f\left(x\right)\) luôn có ít nhất 2 nghiệm

TH3: \(m< -1\) tương tự ta có: \(\lim\limits_{x\rightarrow+\infty}f\left(x\right)=\lim\limits_{x\rightarrow-\infty}=+\infty.\left(m+1\right)=-\infty\)

\(\Rightarrow\) Luôn tồn tại 1 giá trị \(x=a>0\) đủ lớn và \(x=b< 0\) đủ nhỏ sao cho \(\left\{{}\begin{matrix}f\left(a\right)< 0\\f\left(b\right)< 0\end{matrix}\right.\)

Lại có \(f\left(-9\right)=9^4-32>0\) \(\Rightarrow\left\{{}\begin{matrix}f\left(-9\right).f\left(a\right)< 0\\f\left(-9\right).f\left(b\right)< 0\end{matrix}\right.\)

\(\Rightarrow\) Pt luôn có ít nhất 2 nghiệm thuộc  \(\left(-\infty;-9\right)\) và \(\left(-9;+\infty\right)\)

Vậy pt luôn có ít nhất 2 nghiệm với mọi m

a) \(\orbr{\begin{cases}x=0\\x=10\end{cases}}\)

b) \(\orbr{\begin{cases}x=2\\x=4\end{cases}}\)

a) 7x(x-10)=0             

<=> 7x=0 hoặc x-10=0

<=>x=0 hoặc x=10

Vậy \(x\in\left\{0;10\right\}\)

b) 17(3x-6)(2x-8)=0

<=>3x-6=0 hoặc 2x-8=0

<=>3x=6 hoặc 2x=8

<=>x=2 hoặc x=4

Vậy \(x\in\left\{2;4\right\}\)

\(a,7x\left(x-10\right)=0\)

\(Th1:7x=0\Leftrightarrow x=0\)

\(Th2:x-10=0\Leftrightarrow x=10\)

\(b,17\left(3x-6\right)\left(2x-8\right)=0\)

\(Th1:17=0\left(vl\right)\)vô nghiệm 

\(Th2:3x-6=0\Leftrightarrow3x=6\Leftrightarrow x=2\)

\(Th3:2x-8=0\Leftrightarrow2x=8\Leftrightarrow x=4\)

a) (x - 12). 105 = 0

    x-12= 0 : 105

    x - 12 = 0

   x = 0+ 12

   x = 12

b) (47.(27 - x) = 94

    27-x =94 : 47

    27 - x = 2

    x = 27 - 2

    x = 25

c) 2x + 69 .2 = 69.5

   2x = 69.5 - 69.2

   2x = 69 . (5 - 2)

   2x = 69.3

   2x = 207

   x = 207 : 2

   x = 103,5

Ta có: \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\Rightarrow\frac{x}{21}=\frac{y}{14}\left(1\right)\)

          \(5y=7z\Rightarrow\frac{y}{7}=\frac{z}{5}\Rightarrow\frac{y}{14}=\frac{z}{10}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x}{63}=\frac{7y}{98}=\frac{5z}{50}=\frac{3x-7y+5z}{63-98+50}=-\frac{30}{15}=-2\)

\(\frac{x}{21}=2\Rightarrow x=42\)

\(\frac{y}{14}=2\Rightarrow y=28\)

\(\frac{z}{10}=2\Rightarrow z=20\)

Vậy x = 42; y = 28; z = 20

`1)(x+2)(x+3)(x-7)(x-8)=144`
`<=>[(x+2)(x-7)][(x+3)(x-8)]=144`
`<=>(x^2-5x-14)(x^2-5x-24)=144`
`<=>(x^2-5x-19)^2-25=144`
`<=>(x^2-5x-19)^2-169=0`
`<=>(x^2-5x-6)(x^2-5x-32)=0`
`+)x^2-5x-6=0`
`<=>` $\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.$
`+)x^2-5x-32=0`
`<=>` $\left[ \begin{array}{l}x=\dfrac{5+3\sqrt{17}}{2}\\x=\dfrac{5-3\sqrt{17}}{2}\end{array} \right.$
Vậy `S={-1,6,\frac{5+3\sqrt{17}}{2},\frac{5-3\sqrt{17}}{2}}`

1: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)=144\)

\(\Leftrightarrow\left(x^2-7x+2x-14\right)\left(x^2-8x+3x-24\right)=144\)

\(\Leftrightarrow\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+336-144=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+192=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-6\left(x^2-5x\right)-32\left(x^2-5x\right)+192=0\)

\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x-6\right)-32\left(x^2-5x-6\right)=0\)

\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x-32\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+1\right)\left(x^2-5x-32\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+1=0\\x^2-5x-32=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\\x=\dfrac{5-3\sqrt{17}}{2}\\x=\dfrac{5+3\sqrt{17}}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{6;-1;\dfrac{5-3\sqrt{17}}{2};\dfrac{5+3\sqrt{17}}{2}\right\}\)

`2)(6x+5)^2(3x+2)(x+1)=35`
`<=>12(6x+5)^2(3x+2)(x+1)=420`
`<=>(6x+5)^2+(6x+4)(6x+6)=420`
Đặt `6x+5=a` 
`pt<=>a^2(a+1)(a-1)=420`
`<=>a^2(a^2-1)-420=0`
`<=>a^4-a^2-420=0`
`<=>` $\left[ \begin{array}{l}a^2=-20(False)\\a^2=21(True)\end{array} \right.$
`<=>` $\left[ \begin{array}{l}a=\sqrt{20}\\a=-\sqrt{20}\end{array} \right.$
`<=>` $\left[ \begin{array}{l}6x+5=\sqrt{20}\\6x+5=-\sqrt{20}\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=\dfrac{\sqrt{20}-5}{6}\\x=\dfrac{-\sqrt{20}-5}{6}\end{array} \right.$
Vậy `S={\frac{\sqrt{20}-5}{6},\frac{-\sqrt{20}-5}{6}}`