Bài này có thật à lmao

Mà ý của bài này là \(\dfrac{x+1}{90}\)+\(\dfrac{x+2}{80}\)+\(\dfrac{x+3}{70}\)+\(\dfrac{x+4}{60}\)phải không🤔

Bài 1: 

a) Ta có: \(\Delta=\left(2m-1\right)^2-4\cdot m\cdot\left(m+2\right)\)

\(\Leftrightarrow\Delta=4m^2-4m+1-4m^2-8m\)

\(\Leftrightarrow\Delta=-12m+1\)

Để phương trình có nghiệm kép thì \(\Delta=0\)

\(\Leftrightarrow-12m+1=0\)

\(\Leftrightarrow-12m=-1\)

hay \(m=\dfrac{1}{12}\)

b) Ta có: \(\Delta=\left(4m+3\right)^2-4\cdot2\cdot\left(2m^2-1\right)\)

\(\Leftrightarrow\Delta=16m^2+24m+9-16m^2+8\)

\(\Leftrightarrow\Delta=24m+17\)

Để phương trình có nghiệm kép thì \(\Delta=0\)

\(\Leftrightarrow24m+17=0\)

\(\Leftrightarrow24m=-17\)

hay \(m=-\dfrac{17}{24}\)

 ĐKXĐ: \(x,y\ne0\)\(\left\{{}\begin{matrix}x+y+\dfrac{1}{x}+\dfrac{1}{y}=4\\x^3+y^3+\dfrac{1}{x^3}+\dfrac{1}{y^3}=4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}+y+\dfrac{1}{y}=4\\\left(x+\dfrac{1}{x}\right)^3+\left(y+\dfrac{1}{y}\right)^3-3\left(x+\dfrac{1}{x}\right)-3\left(y+\dfrac{1}{y}\right)=4\end{matrix}\right.\)

Đặt \(x+\dfrac{1}{x}=a;y+\dfrac{1}{y}=b\left(a,b\ne0\right)\)

\(\Rightarrow hpt\) trở thành:

\(\left\{{}\begin{matrix}a+b=4\left(1\right)\\a^3+b^3-3a-3b=4\left(2\right)\end{matrix}\right.\) 

Từ (1) \(\Rightarrow a=4-b\) Thay vào (2) ta được:

\(\left(4-b\right)^3+b^3-3\left(4-b\right)-3b=4\Leftrightarrow64-48b+12b^2-b^3+b^3-12+3b-3b-4=0\Leftrightarrow12b^2-48b+60=0\Leftrightarrow b^2-4b+5=0\Leftrightarrow b^2-4b+4+1=0\Leftrightarrow\left(b-2\right)^2+1=0\) Vô lí \(\Rightarrow\) ko có a,b \(\Rightarrow\) ko có x,y

Vậy hpt vô nghiệm

\(x^2-4=2\left(x-2\right)\left(x+3\right)\)

\(\Leftrightarrow x^2-4=2\left(x^2+3x-2x-6\right)\)

\(\Leftrightarrow x^2-4=2x^2+2x-12\)

\(\Leftrightarrow x^2-2x^2-2x=-12+4\)

\(\Leftrightarrow-x^2-2x=-8\)

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

\(\Leftrightarrow-x^2+2x-4x+8=0\)

\(\Leftrightarrow-x\left(x-2\right)-4\left(x-2\right)=0\)

\(\Leftrightarrow\left(-x-4\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x-4=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{-4;2\right\}\)

\(x^2-4=2\left(x-2\right)\left(x+3\right)\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)=2\left(x-2\right)\left(x+3\right)\)

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

\(\Leftrightarrow\left(x-2\right)\left[\left(x+2\right)-2\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-2x-6\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

$(x^2+1)(x-1)=0$

$\Rightarrow x^2+1=0$ hoặc $x-1=0$

$\Rightarrow x=1$

Vậy $x=1$.

$x^3+1=x^2+x$

$\Rightarrow x^3-x^2-x+1=0$

$\Rightarrow x^2(x-1)-(x-1)=0$

$\Rightarrow(x-1)(x^2-1)=0$

$\Rightarrow(x-1)^2(x+1)=0$

$\Rightarrow x=1$ hoặc $x=-1$.

Vậy $x\in{-1;1}$.

1: \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)

=>\(\frac{4\left(3x-2\right)}{12}-\frac{24}{12}=\frac{3\left(4x+1\right)}{12}\)

=>4(3x-2)-24=3(4x+1)

=>12x-8-24=12x+3

=>-32=3(vô lý)

=>x∈∅

2: \(\frac{x-3}{4}+\frac{2x-1}{3}=\frac{2-x}{6}\)

=>\(\frac{3\left(x-3\right)+4\left(2x-1\right)}{12}=\frac{2\left(2-x\right)}{12}\)

=>3(x-3)+4(2x-1)=2(2-x)

=>3x-9+8x-4=4-2x

=>11x-13=4-2x

=>13x=17

=>\(x=\frac{17}{13}\)

3: \(\frac12\left(x+1\right)+\frac14\left(x+3\right)=3-\frac13\left(x+2\right)\)

=>\(\frac12x+\frac12+\frac14x+\frac34=3-\frac13x-\frac23\)

=>\(\frac34x+\frac54=-\frac13x+\frac73\)

=>\(\frac34x+\frac13x=\frac73-\frac54\)

=>\(\frac{13}{12}x=\frac{28-15}{12}=\frac{13}{12}\)

=>x=1

4: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

=>\(\frac{6\left(x+4\right)}{30}+\frac{30\left(-x+4\right)}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)

=>6(x+4)+30(-x+4)=10x-15(x-2)

=>6x+24-30x+120=10x-15x+30

=>-24x+144=-5x+30

=>-19x=-114

=>x=6

5: \(\frac{4-5x}{6}=2\cdot\frac{\left(-x+1\right)}{2}\)

=>\(\frac{-5x+4}{6}=-x+1\)

=>-5x+4=6(-x+1)

=>-5x+4=-6x+6

=>-5x+6x=6-4

=>x=2

1: \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)

=>\(\frac{4\left(3x-2\right)}{12}-\frac{24}{12}=\frac{3\left(4x+1\right)}{12}\)

=>4(3x-2)-24=3(4x+1)

=>12x-8-24=12x+3

=>-32=3(vô lý)

=>x∈∅

2: \(\frac{x-3}{4}+\frac{2x-1}{3}=\frac{2-x}{6}\)

=>\(\frac{3\left(x-3\right)+4\left(2x-1\right)}{12}=\frac{2\left(2-x\right)}{12}\)

=>3(x-3)+4(2x-1)=2(2-x)

=>3x-9+8x-4=4-2x

=>11x-13=4-2x

=>13x=17

=>\(x=\frac{17}{13}\)

3: \(\frac12\left(x+1\right)+\frac14\left(x+3\right)=3-\frac13\left(x+2\right)\)

=>\(\frac12x+\frac12+\frac14x+\frac34=3-\frac13x-\frac23\)

=>\(\frac34x+\frac54=-\frac13x+\frac73\)

=>\(\frac34x+\frac13x=\frac73-\frac54\)

=>\(\frac{13}{12}x=\frac{28-15}{12}=\frac{13}{12}\)

=>x=1

4: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

=>\(\frac{6\left(x+4\right)}{30}+\frac{30\left(-x+4\right)}{30}=\frac{10x}{30}-\frac{15\left(x-2\right)}{30}\)

=>6(x+4)+30(-x+4)=10x-15(x-2)

=>6x+24-30x+120=10x-15x+30

=>-24x+144=-5x+30

=>-19x=-114

=>x=6

5: \(\frac{4-5x}{6}=2\cdot\frac{\left(-x+1\right)}{2}\)

=>\(\frac{-5x+4}{6}=-x+1\)

=>-5x+4=6(-x+1)

=>-5x+4=-6x+6

=>-5x+6x=6-4

=>x=2

a) \(\frac{1}{x-1}\)+\(\frac{2}{x+1}\)=\(\frac{x}{x^2-1}\) (ĐKXĐ:x≠1;x≠-1)

⇔\(\frac{x+1}{\left(x-1\right)\left(x+1\right)}\)+\(\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)=\(\frac{x}{\left(x-1\right)\left(x+1\right)}\)

⇒x+1+2x-2=x

⇔2x-1=0

⇔x=\(\frac{1}{2}\) (TMĐKXĐ)

Vậy tập nghiệm của phương trình đã cho là:S={\(\frac{1}{2}\)}

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