\(\frac{x+29}{31}+\frac{x+27}{33}=\frac{x+17}{43}+\frac{x+15}{45}\)

\(\frac{x+29}{31}+1+\frac{x+27}{33}+1=\frac{x+17}{43}+1+\frac{x+15}{45}+1\)

\(\frac{x+60}{31}+\frac{x+60}{33}=\frac{x+60}{43}+\frac{x+60}{45}\)

\(\left(x+60\right)\left(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\right)=0\)

VÌ \(\frac{1}{31}+\frac{1}{33}-\frac{1}{43}-\frac{1}{45}\ne0\)

\(\Rightarrow x+60=0\)

\(\Rightarrow x=-60\)

Câu hỏi của honoka sonoka - Toán lớp 6 - Học toán với OnlineMath

ket qua x=-60

Bài làm

\(\frac{x+19}{27}-\frac{x+17}{29}=\frac{x+15}{31}-\frac{x+13}{33}\)

\(\Leftrightarrow\left(\frac{x+19}{27}+1\right)-\left(\frac{x+17}{29}+1\right)=\left(\frac{x+15}{31}+1\right)-\left(\frac{x+13}{33}+1\right)\)

\(\Leftrightarrow\frac{x+46}{27}-\frac{x+46}{29}=\frac{x+46}{31}-\frac{x+46}{33}\)

\(\Leftrightarrow\left(x+46\right).\frac{1}{27}-\left(x+46\right).\frac{1}{29}=\left(x+46\right).\frac{1}{31}-\left(x+46\right).\frac{1}{33}\)

\(\Leftrightarrow\left(x+46\right).\frac{1}{27}-\left(x+46\right).\frac{1}{29}-\left(x+46\right).\frac{1}{31}+\left(x+46\right).\frac{1}{33}=0\)

\(\Leftrightarrow\left(x+46\right)\left(\frac{1}{27}-\frac{1}{29}-\frac{1}{31}\right)=0\)

Mà \(\left(\frac{1}{27}-\frac{1}{29}-\frac{1}{31}\right)>0\forall x\)

\(\Leftrightarrow x+46=0\)

\(\Leftrightarrow x=-46\)

Vậy phương trình trên có tập nghiệm S = { -46 }

# Học tốt #

b) Đặt x² + x + 1 = t > 0 (dễ c/m t > 0 rồi ha)

Khi đó, pt tương đương: \(t\left(t+1\right)=12\Leftrightarrow t^2+t-12=0\Leftrightarrow\left[{}\begin{matrix}t=3\\t=-4\left(L\right)\end{matrix}\right.\)

t = 3 suy ra \(x^2+x+1=3\Leftrightarrow x^2+x-2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy...

c) Chị xem lại đề giúp em ạ.

1) \(\left|x-2\right|+2=x\)

\(\Leftrightarrow\left|x-2\right|=x-2\)

\(\Leftrightarrow x-2\ge0\Leftrightarrow x\ge2\)

2) \(x^2+5x+4=0\)

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

\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)

3) \(8\sqrt{x}=x^2\)

Bình phương hai vế, ta được: \(64x=x^4\)

\(\Leftrightarrow x^4-64x=0\)

\(\Leftrightarrow x\left(x^3-64\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^3-64=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

4) \(\frac{x+29}{31}-\frac{x+27}{33}=\frac{x+17}{43}-\frac{x+15}{45}\)

\(\Leftrightarrow\frac{x+29}{31}-\frac{x+27}{33}-\frac{x+17}{43}+\frac{x+15}{45}=0\)

\(\Leftrightarrow\frac{x+29}{31}+1-\frac{x+27}{33}-1-\frac{x+17}{43}-1+\frac{x+15}{45}+1=0\)

\(\Leftrightarrow\frac{x+60}{31}+\frac{x+60}{45}-\frac{x+60}{33}-\frac{x+60}{43}=0\)

\(\Leftrightarrow\left(x+60\right)\left(\frac{1}{31}+\frac{1}{45}-\frac{1}{33}-\frac{1}{43}\right)=0\)

\(\Leftrightarrow x+60=0\Leftrightarrow x=-60\)

5)\(\left|x-1\right|+3x=1\)

\(\Leftrightarrow\left|x-1\right|=1-3x\)(1)

* Nếu \(x\ge1\)thì \(\left(1\right)\Leftrightarrow x-1=1-3x\Leftrightarrow4x=2\Leftrightarrow x=\frac{1}{2}\left(L\right)\)

* Nếu \(x< 1\)thì \(\left(1\right)\Leftrightarrow1-x=1-3x\Leftrightarrow2x=0\Leftrightarrow x=0\left(TM\right)\)

Vậy x = 0