giúp mình với

x=−1−√13/2,−1+√13/2

ĐKXĐ:...

a. Đặt \(\left\{{}\begin{matrix}\sqrt{2x^2+4x+16}=a>0\\\sqrt{x+70}=b\ge0\end{matrix}\right.\)

\(\Rightarrow6x^2+10x-92=3a^2-2b^2\)

Pt trở thành:

\(3a^2-2b^2+ab=0\)

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

\(\Leftrightarrow3a=2b\)

\(\Leftrightarrow9\left(2x^2+4x+16\right)=4\left(x+70\right)\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)

Phương trình trở thành:

\(a^2+2+ab=3a+b\)

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

\(\Leftrightarrow\left(a-1\right)\left(a-2\right)+b\left(a-1\right)=0\)

\(\Leftrightarrow\left(a-1\right)\left(a+b-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{x+1}+\sqrt{1-x}=2\end{matrix}\right.\)

\(\Leftrightarrow...\)

\(Đk:x\ge\dfrac{3}{2}\Rightarrow x>0\)

\(x^3-4x^2+5x-1-\sqrt{2x-3}=0\)

\(\Leftrightarrow2x^3-8x^2+10x-2-2\sqrt{2x-3}=0\)

\(\Leftrightarrow\left(2x^3-8x^2+8x\right)+\left[\left(2x-3\right)-2\sqrt{2x-3}+1\right]=0\)

\(\Leftrightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2=0\)

Ta có: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2\ge0\left(x>0\right)\\\left(\sqrt{2x-3}-1\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow2x\left(x-2\right)^2+\left(\sqrt{2x-3}-1\right)^2\ge0\)

Do đó: \(\left\{{}\begin{matrix}2x\left(x-2\right)^2=0\\\left(\sqrt{2x-3}-1\right)^2=0\end{matrix}\right.\Leftrightarrow x=2\)

Thử lại ta có x=2 là nghiệm duy nhất của phương trình đã cho.

x^3-4x^2+5x-1-căn 2x-3=0

=>\(x^3-4x^2+5x-2-\left(\sqrt{2x-3}-1\right)=0\)

=>\(\left(x-1\right)\left(x-2\right)^2-\dfrac{2x-3-1}{\sqrt{2x-3}+1}=0\)

=>\(\left(x-2\right)\left[\left(x-1\right)\left(x-2\right)-\dfrac{2}{\sqrt{2x-3}+1}\right]=0\)

=>x-2=0

=>x=2

\(\dfrac{x}{2}\left(4x-3\right)+2\left(3-x\right)\left(x+4\right)\le0\)

\(\Leftrightarrow\dfrac{4x^2}{2}-\dfrac{3x}{2}+2\left(3x+12-x^2-4x\right)\le0\)

\(\Leftrightarrow\dfrac{4x^2-3x}{2}+6x+24-2x^2-8x\le0\)

\(\Leftrightarrow\dfrac{4x^2-3x+2\left(6x+24-2x^2-8x\right)}{2}\le0\)

\(\Leftrightarrow4x^2-3x+12x+48-4x^2-16x\le0\)

\(\Leftrightarrow-7x\le-48\)

\(\Leftrightarrow x\ge\dfrac{48}{7}\)

=>-7x+48≤0

<=>-7x≤-48

<=>(-7x)(-1)≥(-48)(-1)

<=>\(\dfrac{7x}{7}\)≥\(\dfrac{48}{7}\)

<=>x≥\(\dfrac{48}{7}\)

Bài 1 bạn tìm quanh quanh đây, mình thấy có bài y hệt rồi nên ko làm nữa

Bài 2 như sau:

ĐKXĐ: \(x\ge\dfrac{-1}{16}\)

\(x^2-x-20-2\left(\sqrt{16x+1}-9\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)-2\dfrac{\left(\sqrt{16x+1}-9\right)\left(\sqrt{16x+1}+9\right)}{\sqrt{16x+1}+9}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4\right)-\dfrac{32\left(x-5\right)}{\sqrt{16x+1}+9}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+4-\dfrac{32}{\sqrt{16x+1}+9}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\Rightarrow x=5\\x+4-\dfrac{32}{\sqrt{16x+1}+9}=0\left(1\right)\end{matrix}\right.\)

Xét phương trình (1): ta có \(x+4\ge-\dfrac{1}{16}+4=\dfrac{63}{16}\) \(\forall x\ge-\dfrac{1}{16}\)

\(\sqrt{16x+1}+9\ge9\Rightarrow\dfrac{32}{\sqrt{16x+1}+9}\le\dfrac{32}{9}\) \(\forall x\ge-\dfrac{1}{16}\)

Mà \(\dfrac{63}{16}-\dfrac{32}{9}=\dfrac{55}{144}>0\) \(\Rightarrow x+4-\dfrac{32}{\sqrt{16x+1}+9}>0\) \(\forall x\ge-\dfrac{1}{16}\)

\(\Rightarrow\) pt (1) vô nghiệm

Vậy pt đã cho có nghiệm duy nhất \(x=5\)

a)Dat \(x^2-4x+3=a;x^2-7x+6=b \Rightarrow a+b=2x^2-11x+9\)

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