Lời giải:
Để pt có 2 nghiê pb thì:

$\Delta'=1-(m-3)>0\Leftrightarrow m< 4$

Áp dụng định lý Viet: \(\left\{\begin{matrix} x_1+x_2=2\\ x_1x_2=m-3\end{matrix}\right.\)

Khi đó:
\(x_1^2-2x_2+x_1x_2=-12\)

\(\Leftrightarrow x_1^2-2(2-x_1)+x_1(2-x_1)=-12\)

\(\Leftrightarrow x_1=-2\Leftrightarrow x_2=2-x_1=4\)

$m-3=x_1x_2=(-2).4=-8$

$\Leftrightarrow m=-5$ (tm)

\(x^2-x+1-m=0\left(1\right)\\ \text{PT có 2 nghiệm }x_1,x_2\\ \Leftrightarrow\Delta=1-4\left(1-m\right)\ge0\\ \Leftrightarrow4m-3\ge0\Leftrightarrow m\ge\dfrac{3}{4}\\ \text{Vi-ét: }\left\{{}\begin{matrix}x_1+x_2=1\\x_1x_2=1-m\end{matrix}\right.\\ \text{Ta có }5\left(\dfrac{1}{x_1}+\dfrac{1}{x_2}\right)-x_1x_2+4=0\\ \Leftrightarrow5\cdot\dfrac{x_1+x_2}{x_1x_2}-x_1x_2+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m-1+4=0\\ \Leftrightarrow\dfrac{5}{1-m}+m+3=0\\ \Leftrightarrow5+\left(1-m\right)\left(m+3\right)=0\\ \Leftrightarrow m^2+2m-8=0\\ \Leftrightarrow m^2-2m+4m-8=0\\ \Leftrightarrow\left(m-2\right)\left(m+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}m=2\left(n\right)\\m=-4\left(l\right)\end{matrix}\right.\)

Vậy $m=2$

3x(2-x)-5=1-(3x²+2)

<=>6x-3x²-5=-3x²-2

<=>6x=3

<=>x=1/2

\(a,\Delta'=\left(-1\right)^2-\left(m-3\right)=1-m+3=4-m\)

Để pt trên có nghiệm thì \(4-m\ge0\Leftrightarrow m\le4\)

b, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=2\\x_1x_2=m-3\end{matrix}\right.\)

\(\left(x_1+x_2\right)^2=16+2x_1x_2\\ \Leftrightarrow2^2=16+2\left(m-3\right)\\ \Leftrightarrow2m-6+16-4=0\\ \Leftrightarrow2m+6=0\\ \Leftrightarrow m=-3\left(tm\right)\)

Bài giải:

x thuộc BC(8,10,154)

8=2^3

10=2.5

154=2.7.11

Vậy BCNN(8,10,154)=2^3.5.7.11=3080

B(3080)={0;3080;6160;...}

Vì 50<x<500 nên x rỗng.

Bài này mình làm quen rồi nên tin mk nhé!

HỌC TỐT NHÉ! 

Ta có:

     (2 - 3x)(x + 8) = (3x - 2)(3 - 5x)

⇔ (2 - 3x)(x + 8) - (3x - 2)(3 - 5x) = 0

⇔ (2 - 3x)(x + 8) + (2 - 3x)(3 - 5x) = 0

⇔ (2 - 3x)(x + 8 + 3 - 5x) = 0

⇔ (2 - 3x)(11 - 4x) = 0

⇔ 2 - 3x = 0 hay 11 - 4x = 0

⇔ 2 = 3x hay 11 = 4x

⇔ x = \(\dfrac{2}{3}\) hay x = \(\dfrac{11}{4}\)

Vậy tập nghiệm của pt S = \(\left\{\dfrac{2}{3};\dfrac{11}{4}\right\}\)

khum bt mới lớp 5 hoi à(k10)

<=> (2-3x ) (x+8) + (2-3x ) (3-5x)=0
<=> (2-3x ) ( x+8 +  3-5x ) =0 
<=> (2-3x ) ( 11 - 4x ) = 0
 => 2-3x  =0 hoặc 11-4x =0  
       3x = 2            4x =11
         x = 2/3         x    = 11/4

`[x-4]/[x+4]-x/[x-4]=[3x-14]/[x^2-16]`         `ĐK: x \ne +-4`

`<=>[(x-4)^2-x(x+4)]/[(x-4)(x+4)]=[3x-14]/[(x-4)(x+4)]`

   `=>x^2-8x+16-x^2-4x=3x-14`

`<=>3x+8x+4x=16+14`

`<=>15x=30`

`<=>x=2` (t/m)

Vậy `S={2}`

`(x - 4)/(x + 4) - x/(x - 4) = (3x - 14)/(x^2 - 16)`

`=>` `x = 2`

\(\Leftrightarrow2cos4x\left(cos2x-sin2x\right)=0\)

\(\Leftrightarrow cos4x=0\) (do \(cos4x=cos^22x-sin^22x\) đã bao hàm \(cos2x-sin2x\))

\(\Rightarrow4x=\dfrac{\pi}{2}+k\pi\)

\(\Rightarrow x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\)

ĐKXĐ: \(x\ge1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[]{x-1}=a\ge0\\\sqrt[3]{2-x}=b\end{matrix}\right.\) \(\Rightarrow a^2+b^3=1\)

Ta được hệ: 

\(\left\{{}\begin{matrix}a+b=1\\a^2+b^3=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=1-a\\a^2+b^3=1\end{matrix}\right.\)

\(\Rightarrow a^2+\left(1-a\right)^3=1\)

\(\Leftrightarrow a^3-4a^2+3a=0\)

\(\Leftrightarrow a\left(a-1\right)\left(a-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt[]{x-1}=0\\\sqrt[]{x-1}=1\\\sqrt[]{x-1}=3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=10\end{matrix}\right.\)