Chương 3: PHƯƠNG TRÌNH, HỆ PHƯƠNG TRÌNH

ta có \(\left\{{}\begin{matrix}\left|4x+7\right|=4x+7\left(khi4x+7\ge0\right)\\\left|4x+7\right|=-4x-7\left(khi4x+7< 0\right)\end{matrix}\right.\)

suy ra ta có : \(\left|4x+7\right|=4x+7\) khi \(4x+7\ge0\) \(\Leftrightarrow4x\ge-7\Leftrightarrow x\ge\dfrac{-7}{4}\)

vậy \(x\ge\dfrac{-7}{4}\)

ta có : \(\left|4x+7\right|=2x+5\Leftrightarrow\left(\left|4x+7\right|\right)^2=\left(2x+5\right)^2\)

\(\Leftrightarrow\left(4x+7\right)^2=\left(2x+5\right)^2\Leftrightarrow16x^2+56x+49=4x^2+20x+25\)

\(\Leftrightarrow16x^2+56x+49-4x^2-20x-25=0\)

\(\Leftrightarrow12x^2+36x+24=0\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

thử lại ta thấy : cả 2 nghiệm này đều thỏa mảng phương trình đầu

vậy \(x=-1;x=-2\)

ta có : \(\left|2x-1\right|=x+3\Leftrightarrow\left(\left|2x-1\right|\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow\left(2x-1\right)^2=\left(x+3\right)^2\Leftrightarrow4x^2-4x+1=x^2+6x+9\)

\(\Leftrightarrow4x^2-4x+1-x^2-6x-9=0\)

\(\Leftrightarrow3x^2-10x-8=0\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{-2}{3}\end{matrix}\right.\)

thử lại ta thấy : cả 2 nghiệm này đều thỏa mảng phương trình đầu

vậy \(x=4;x=\dfrac{-2}{3}\)

a: \(x^2-2x+\left|x-1\right|-1=0\)

\(\Leftrightarrow x^2-2x+1+\left|x-1\right|-2=0\)

\(\Leftrightarrow\left(\left|x-1\right|\right)^2+\left|x-1\right|-2=0\)

\(\Leftrightarrow\left(\left|x-1\right|+2\right)\left(\left|x-1\right|-1\right)=0\)

=>|x-1|=1

=>x-1=1 hoặc x-1=-1

=>x=2 hoặc x=0

b: \(4x^2-4x-\left|2x-1\right|-1=0\)

\(\Leftrightarrow4x^2-4x+1-\left|2x-1\right|-2=0\)

\(\Leftrightarrow\left(\left|2x-1\right|\right)^2-\left|2x-1\right|-2=0\)

\(\Leftrightarrow\left(\left|2x-1\right|-2\right)\left(\left|2x-1\right|+1\right)=0\)

=>|2x-1|=2

=>2x-1=2 hoặc 2x-1=-2

=>x=3/2 hoặc x=-1/2

c: \(\left|2x-5\right|+\left|2x^2-7x+5\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\\left(2x-5\right)\left(x-1\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{5}{2}\)

d: \(x^2-2x-5\left|x-1\right|-5=0\)

\(\Leftrightarrow x^2-2x+1-5\left|x-1\right|-6=0\)

\(\Leftrightarrow\left(\left|x-1\right|\right)^2-5\left|x-1\right|-6=0\)

\(\Leftrightarrow\left(\left|x-1\right|-6\right)\left(\left|x-1\right|+1\right)=0\)

=>|x-1|=6

=>x-1=6 hoặc x-1=-6

=>x=7 hoặc x=-5

a. \(\left|x^2-4x-5\right|=4x-17\)

\(\Leftrightarrow\left\{{}\begin{matrix}4x-17\ge0\\\left[{}\begin{matrix}x^2-4x-5=4x-17\\x^2-4x-5=17-4x\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{17}{4}\\\left[{}\begin{matrix}\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\\\left[{}\begin{matrix}x=\sqrt{22}\\x=-\sqrt{22}\end{matrix}\right.\end{matrix}\right.\end{matrix}\right.\)

Vậy tập ngiệm của pt trên là: \(S=\left\{\sqrt{22};6\right\}\)

a, ĐKXĐ: \(-3\le x\le6\)

\(pt\Leftrightarrow3+x+6-x+2\sqrt{\left(3+x\right)\left(6-x\right)}=9\)

\(\Leftrightarrow\sqrt{\left(3+x\right)\left(6-x\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

b, ĐKXĐ: \(x\ge4\)

\(pt\Leftrightarrow\sqrt{x-4+4\sqrt{x-4}+4}+x+2+\sqrt{x-4}=8\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}+x+2+\sqrt{x-4}=8\)

\(\Leftrightarrow\sqrt{x-4}+2+x+2+\sqrt{x-4}=8\)

\(\Leftrightarrow2\sqrt{x-4}=4-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}4-x\ge0\\4\left(x-4\right)=\left(4-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\x^2-12x+32=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)

e, Đặt \(y=x-1\) ta có

\(pt\Leftrightarrow\left(y+4\right)^4+\left(y-4\right)^4=1312\)

\(\Leftrightarrow2y^4+192y^2-800=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y^2=4\\y^2=-100\left(l\right)\end{matrix}\right.\Leftrightarrow y=\pm2\)

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

mũ 4 hay nhan 4 vay

ta có:

X⁴ z⁴ y⁴  luôn>0

x-y>=\(\sqrt{2xy}\)  >0

tương tự z-x,  y-z  =>A luôn dương