sao k ai trả lời zậy ta

Nhấn máy tính: 

+ giải hpt x²-4x+3: mode=> 5:EQN=> số 3=> 1=> = => -4 => = => 3=> X₁=3 => = => X₂=1

=> Thay vào=> Đưa vô căn bậc 2.

+ giải hpt 2x² -3x+1 tương tự như trên.

=> Sau đó thay vô. tính ra

Xin lỗi mình chỉ biết nhiêu đây. lớp 7. Thông cảm.

ĐKXĐ: ...

\(\Leftrightarrow3x-1-x\sqrt{3x-1}+x\sqrt{x+1}-\sqrt{\left(x+1\right)\left(3x-1\right)}=0\)

\(\Leftrightarrow\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)-\sqrt{x+1}\left(\sqrt{3x-1}-x\right)=0\)

\(\Leftrightarrow\left(\sqrt{3x-1}-\sqrt{x+1}\right)\left(\sqrt{3x-1}-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x-1}=\sqrt{x+1}\\\sqrt{3x-1}=x\end{matrix}\right.\)

\(\Leftrightarrow...\)

ĐKXĐ: x \(\ge\)\(\dfrac{1}{3}\)

pt\(\Leftrightarrow\)x(\(\sqrt{x+1}-\sqrt{3x-1}\))+\(\sqrt{3x-1}\left(\sqrt{3x-1}-\sqrt{x+1}\right)\)=0

  \(\Leftrightarrow\)(\(\sqrt{x+1}-\sqrt{3x-1}\))(1-\(\sqrt{3x-1}\))=0

  \(\Leftrightarrow\)\(\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{3x-1}\\1=\sqrt{3x-1}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{2}{3}\end{matrix}\right.\)(t/m x \(\ge\)\(\dfrac{1}{3}\))

Vậy.....................

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

ta có:\(x\left(3-\sqrt{3x-1}\right)\)

=\(3x-x\sqrt{3x-1}\)

=\(3x-1-x\sqrt{3x-1}+1\)

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

Ta có \(\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)

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

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

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

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

⇔\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)+1\)=\(\sqrt{x+1}\left(\sqrt{3x-1}-x\right)+1\)

⇔\(\sqrt{3x-1}\left(\sqrt{3x-1}-x\right)=\sqrt{x+1}\left(\sqrt{3x-1}-x\right)\)

⇔​​\(\sqrt{3x-1}=\sqrt{x+1}\)

⇔​\(3x-1=x+1\)

⇔\(2x=2\)

⇔x=1(N)

​Vậy x=1

ĐKXĐ: \(\frac{2}{3}\le x\le5\)

\(\Leftrightarrow\sqrt{2x+7}\ge\sqrt{5-x}+\sqrt{3x-2}\)

\(\Leftrightarrow2x+7\ge2x+3+2\sqrt{-3x^2+17x-10}\)

\(\Leftrightarrow\sqrt{-3x^2+17x-10}\le2\)

\(\Leftrightarrow-3x^2+17x-10\le4\)

\(\Leftrightarrow3x^2-17x+14\ge0\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge\frac{14}{3}\end{matrix}\right.\)

Kết hợp ĐKXĐ: \(\Rightarrow\left[{}\begin{matrix}\frac{2}{3}\le x\le1\\\frac{14}{3}\le x\le5\end{matrix}\right.\)

Ok here we go

Chắc là ko cần hiểu đâu, nhưng toàn bộ nằm trong quy tắc cơ bản mà: \(\int\left(uv\right)'dx=\int u'vdx+\int uv'dx\)

\(\int f'\left(x\right)dx=f\left(x\right)\) nên \(\int\left(uv\right)'dx=uv\)

\(v'dx=dv\) ; \(u'dx=du\)

ráp vào là thành công thức kia

\(\Leftrightarrow\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=4k+1\)

\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{4k+1}{\sqrt{2}}\)

\(\Rightarrow-1\le\frac{4k+1}{\sqrt{2}}\le1\)

\(\Rightarrow\frac{-\sqrt{2}-1}{4}\le k\le\frac{\sqrt{2}-1}{4}\)

\(\Rightarrow k=0\)

\(\Rightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)