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Câu hỏi của Nguyễn Thùy Chi - Toán lớp 10 | Học trực tuyến

ĐKXĐ: ...

\(\Leftrightarrow\left(x-1\right)\left(x+3-\sqrt{14x-15}\right)-\sqrt{10x-19}+1=0\)

\(\Leftrightarrow x^2+2x-2-\left(x-1\right)\sqrt{14x-15}-\sqrt{10x-19}=0\)

\(\Leftrightarrow x-\sqrt{10x-19}+\left(x-1\right)\left(x+2\right)-\left(x-1\right)\sqrt{14x-15}=0\)

\(\Leftrightarrow\frac{x^2-10x+19}{x+\sqrt{10x-19}}+\left(x-1\right)\left(\frac{x^2-10x+19}{x+2+\sqrt{14x+15}}\right)=0\)

\(\Leftrightarrow\left(x^2-10x+19\right)\left(\frac{1}{x+\sqrt{10x-19}}+\frac{x-1}{x+2+\sqrt{14x+15}}\right)=0\)

\(\Leftrightarrow x^2-10x+19=0\)

câu a

Học tại nhà - Toán - Bài 110035

b,  ĐK \(x\ge-4\)

PT 

<=> \(\left(x-\sqrt{x+4}\right)+\left(\sqrt{2x^2-10x+17}-2x+3\right)=0\)

<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}+\frac{-2x^2+2x+8}{\sqrt{2x^2-10x+17}+2x-3}=0\)với \(x+\sqrt{x+4}\ne0\)

<=> \(\frac{x^2-x-4}{x+\sqrt{x+4}}-\frac{2\left(x^2-x-4\right)}{\sqrt{2x^2-10x+17}+2x-3}=0\)

<=> \(\orbr{\begin{cases}x^2-x-4=0\\\frac{1}{x+\sqrt{x+4}}-\frac{2}{\sqrt{2x^2-10x+17}+2x-3}=0\left(2\right)\end{cases}}\)

Giải (2)

=> \(2x+2\sqrt{x+4}=2x-3+\sqrt{2x^2-10x+17}\)

<=> \(\sqrt{2x^2-10x+17}=2\sqrt{x+4}+3\)

<=> \(2x^2-10x+17=4\left(x+4\right)+9+12\sqrt{x+4}\)

<=> \(x^2-7x-4=6\sqrt{x+4}\)

<=> \(\left(x-6\right)^2+5x-40=6\sqrt{6\left(x-6\right)-5x+40}\)

Đặt x-6=a;\(\sqrt{6\left(x-6\right)-5x+40}=b\)

=> \(\hept{\begin{cases}a^2+5x-40=6b\\b^2+5x-40=6a\end{cases}}\)

=> \(a^2-b^2+6\left(a-b\right)=0\)

<=> \(\orbr{\begin{cases}a=b\\a+b+6=0\end{cases}}\)

+ a=b

=> \(x-6=\sqrt{x+4}\)

=> \(\hept{\begin{cases}x\ge6\\x^2-13x+32=0\end{cases}}\)=> \(x=\frac{13+\sqrt{41}}{2}\)

+ a+b+6=0

=> \(x+\sqrt{x+4}=0\)(loại)

Vậy \(S=\left\{\frac{13+\sqrt{41}}{2};\frac{1+\sqrt{17}}{2}\right\}\)

ĐKXĐ: \(x\ge\frac{9}{10}\)

\(\Leftrightarrow x^2+4x+1-x\sqrt{14x-1}-\sqrt{10x-9}=0\)

\(\Leftrightarrow x\left(x+3-\sqrt{14x-1}\right)+x+1-\sqrt{10x-9}=0\)

\(\Leftrightarrow\frac{x\left[\left(x+3\right)^2-\left(14x-1\right)\right]}{x+3+\sqrt{14x-1}}+\frac{\left(x+1\right)^2-\left(10x-9\right)}{x+1+\sqrt{10x-9}}=0\)

\(\Leftrightarrow\frac{x\left(x^2-8x+10\right)}{x+3+\sqrt{14x-1}}+\frac{x^2-8x+10}{x+1+\sqrt{10x-9}}=0\)

\(\Leftrightarrow\left(x^2-8x+10\right)\left(\frac{x}{x+3+\sqrt{14x-1}}+\frac{1}{x+1+\sqrt{10x-9}}\right)=0\)

\(\Leftrightarrow x^2-8x+10=0\) (casio)

a) ĐK : \(x\ge1\)

pt <=> \(\sqrt{3^2\left(x-1\right)}-\frac{1}{2}\sqrt{2^2\left(x-1\right)}=2\)

<=> \(\left|3\right|\sqrt{x-1}-\frac{1}{2}\cdot\left|2\right|\sqrt{x-1}=2\)

<=> \(3\sqrt{x-1}-1\sqrt{x-1}=2\)

<=> \(2\sqrt{x-1}=2\)

<=> \(\sqrt{x-1}=1\)

<=> \(x-1=1\)=> \(x=2\)( tm )

b) \(3x-\sqrt{49-14x+x^2}=15\)

<=> \(\sqrt{x^2-14x+49}=3x-15\)

<=> \(\sqrt{\left(x-7\right)^2}=3x-15\)

<=> \(\left|x-7\right|=3x-15\)(1)

Với x < 7

(1) <=> 7 - x = 3x - 15

     <=> -x - 3x = -15 - 7

     <=> -4x = -22

     <=> x = 11/2 ( tm )

Với x ≥ 7

(1) <=> x - 7 = 3x - 15

      <=> x - 3x = -15 + 7

      <=> -2x = -8

      <=> x = 4 ( ktm )

Vậy x = 11/2

a) \(ĐKXĐ:x\ge1\)

\(\sqrt{9x-9}-\frac{1}{2}\sqrt{4x-4}=2\)

\(\Leftrightarrow\sqrt{9.\left(x-1\right)}-\frac{1}{2}.\sqrt{4\left(x-1\right)}=2\)

\(\Leftrightarrow3\sqrt{x-1}-\frac{1}{2}.2\sqrt{x-1}=2\)

\(\Leftrightarrow3\sqrt{x-1}-\sqrt{x-1}=2\)

\(\Leftrightarrow2\sqrt{x-1}=2\)

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

\(\Leftrightarrow x-1=1\)\(\Leftrightarrow x=2\)( thỏa mãn ĐKXĐ )

Vậy phương trình có nghiệm là \(x=2\)

b) \(3x-\sqrt{49-14x+x^2}=15\)

\(\Leftrightarrow3x-\sqrt{\left(7-x\right)^2}=15\)

\(\Leftrightarrow3x-\left|7-x\right|=15\)

+) TH1: Nếu \(7-x< 0\)\(\Leftrightarrow x>7\)

thì \(3x-\left(x-7\right)=15\)

\(\Leftrightarrow3x-x+7=15\)\(\Leftrightarrow2x=8\)

\(\Leftrightarrow x=4\)( không thỏa mãn )

+) TH2: Nếu \(7-x\ge0\)\(\Leftrightarrow x\le7\)

thì \(3x-\left(7-x\right)=15\)

\(\Leftrightarrow3x-7+x=15\)

\(\Leftrightarrow4x=22\)\(\Leftrightarrow x=\frac{22}{4}\)( thỏa mãn ĐKXĐ )

Vậy nghiệm của phương trình là \(x=\frac{22}{4}\)