liên hợp nghiệm kép nha

(1) \(\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)

cái này đâu ra z ???

nguyen van tuan: hì, xin lỗi, làm hơi tắt ^^!

\(\left(1\right)\Leftrightarrow\left(x+1\right)\sqrt{16x+17}=\left(x+1\right)\left(x-\dfrac{23}{8}\right)\Leftrightarrow\left(x+1\right)\sqrt{16x+17}-\left(x+1\right)\left(x-\dfrac{23}{8}\right)=0\Leftrightarrow\left(x+1\right)\left(\sqrt{16x+17}-x+\dfrac{23}{8}\right)=0\)

b: \(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{23}{8}\\64x^2-368x+529-16x-17=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{23}{8}\\64x^2-384x+512=0\end{matrix}\right.\Leftrightarrow x=4\)

c: \(\Leftrightarrow3\sqrt{x^2+3x}=\left(2x-x^2+10-5x\right)\)

\(\Leftrightarrow3\sqrt{x^2+3x}=-\left(x^2+3x-10\right)\)

Đặt \(\sqrt{x^2+3x}=a\)

Ta có: \(3a=-\left(a^2-10\right)=-a^2+10\)

\(\Leftrightarrow a^2+3a-10=0\)

=>(a+5)(a-2)=0

=>a=2

=>x²+3x=4

=>(x+4)(x-1)=0

=>x=1 hoặc x=-4

Sửa đề : \(x^2+3=..\) nhé

a) \(\sqrt{1-8x+16x^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{1^2-2\cdot4x\cdot1+\left(4x\right)^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{\left(4x-1\right)^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\left|4x-1\right|=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1=\dfrac{1}{3}\left(ĐK:x\ge\dfrac{1}{4}\right)\\4x-1=\dfrac{1}{3}\left(ĐK:x< \dfrac{1}{4}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=\dfrac{1}{6}\left(tm\right)\end{matrix}\right.\)

b) \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\) (ĐK: \(x\ge2\)) 

\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)

\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)

\(\Leftrightarrow6\sqrt{x-2}=18\)

\(\Leftrightarrow\sqrt{x-2}=3\)

\(\Leftrightarrow x-2=9\)

\(\Leftrightarrow x=9+2\)

\(\Leftrightarrow x=11\left(tm\right)\)

\(1,\dfrac{x-1}{3}=x+1\\ \Leftrightarrow x-1=3x+3\\ \Leftrightarrow3x-x=3+1\\ \Leftrightarrow x=2\)

PT có tập nghiệm S = {2}

\(2,\sqrt{16x^2+8x+1}-2=x\\ \Leftrightarrow\sqrt{\left(4x+1\right)^2}-2=x\\\Leftrightarrow 4x+1-2=x\\ \Leftrightarrow4x-x=2-1\\ \Leftrightarrow x=\dfrac{1}{3}\)

PT có tập nghiệm S = {1/3}

\(3,\left\{{}\begin{matrix}2x+y=17\\x-2y=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x+y=17\\2x-4y=2\end{matrix}\right.\\ \Leftrightarrow\left(2x+y\right)-\left(2x-4y\right)=17-2\\ \Leftrightarrow5y=15\\ \Leftrightarrow y=3\\ \Leftrightarrow2x+3=17\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\)

PTHH có tập nghiệm (x; y) là (7; 3)

\(\sqrt[]{8x^2-16x+10}+\sqrt[]{2x^2-4x+10}=\sqrt[]{7-x^2+2x}\)

\(\Leftrightarrow\sqrt[]{8x^2-16x+10}=\dfrac{1}{4}\sqrt[]{2\left(7-x^2+2x\right)}-\sqrt[]{2x^2-4x+10}\)

\(\Leftrightarrow\sqrt[]{8x^2-16x+10}=\dfrac{1}{4}\sqrt[]{14-2x^2+4x}-\sqrt[]{2x^2-4x+10}\left(1\right)\)

Áp dụng BĐT Bunhiacopxki ta được:

\(\left[\dfrac{1}{4}\sqrt[]{14-2x^2+4x}+\left(-1\right).\sqrt[]{2x^2-4x+10}\right]^2\le\left(\dfrac{1}{16}+1\right)\left(14-2x^2+4x+2x^2-4x+10\right)=\dfrac{17}{16}.24=\dfrac{51}{2}\)

Dấu "=" xảy ra khi và chỉ khi

\(\sqrt[]{14-2x^2+4x}+4\sqrt[]{2x^2-4x+10}=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}14-2x^2+4x=0\\2x^2-4x+10=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}14+2-2\left(x^2-2x+1\right)=0\\2\left(x^2-2x+1\right)+10-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\left(x-1\right)^2+16=0\\2\left(x-1\right)^2+8=0\end{matrix}\right.\) \(\Leftrightarrow x\in\varnothing\)

\(pt\left(1\right)\Leftrightarrow8x^2-16x+10=\dfrac{51}{2}\)

\(\Leftrightarrow16x^2-32x+20-51=0\)

\(\Leftrightarrow16x^2-32x-31=0\left(2\right)\)

\(\Delta'=256+496=752>0\)

\(\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{47}\)

\(pt\left(2\right)\) có 2 nghiệm phân biệt

\(x=\dfrac{16\pm4\sqrt[]{47}}{16}=\dfrac{4\pm\sqrt[]{47}}{4}\)

Cách giải trên đã sai, mình giải lại

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

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

Vì \(\left(x-1\right)^2\ge0,\forall x\in R\)

\(\Rightarrow\left\{{}\begin{matrix}8\left(x-1\right)^2+2\ge2,\forall x\in R\\2\left(x-1\right)^2+2\ge2,\forall x\in R\\8-\left(x-1\right)^2\le8,\forall x\in R\end{matrix}\right.\)

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

Thay \(x=1\) vào \(\left(2\right)\) ta được

\(\sqrt[]{8.0+2}+\sqrt[]{2.0+2}=\sqrt[]{8-0}\)

\(\Leftrightarrow\sqrt[]{2}+\sqrt[]{2}=\sqrt[]{8}=2\sqrt[]{2}\left(đúng\right)\)

Vậy nghiệm của phương trình đã cho là \(x=1\)

\(a=\sqrt{25x^2-10x+1+16}=\sqrt{\left(5x-1\right)^2+16}\ge\sqrt{16}=4\)

\(a_{min}=4\) khi \(5x-1=0\Leftrightarrow x=\frac{1}{5}\)

\(b=\sqrt{x^2-10x+25+5}=\sqrt{\left(x-5\right)^2+5}\ge\sqrt{5}\)

\(b_{min}=\sqrt{5}\) khi \(x=5\)

\(c=\sqrt{-16x^2-8x-1+4}=\sqrt{4-\left(4x+1\right)^2}\le\sqrt{4}=2\)

\(c_{max}=2\) khi \(x=-\frac{1}{4}\)