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a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
ĐK \(x\le\frac{-5-\sqrt{41}}{8}\)hoặc \(x\ge\frac{1+\sqrt{5}}{2}\)
Nhân liên hợp 2 vế ta có:
=> \(\left(4x^2+5x-1-4x^2+4x+4\right)=3\left(3x+1\right)\left(\sqrt{4x^2+5x-1}+2\sqrt{x^2-x-1}\right)\)<=> \(3\left(3x+1\right)=3\left(3x+1\right)\left(\sqrt{4x^2+5x-1}+2\sqrt{x^2-x-1}\right)\)
<=>\(\left[{}\begin{matrix}x=-\frac{1}{3}\left(koTMĐKXĐ\right)\\\sqrt{4x^2+5x-1}+2\sqrt{x^2-x-1}=1\left(2\right)\end{matrix}\right.\)
Kết hợp (2) với PT ban đầu ta có:
=> \(2\sqrt{4x^2+5x-1}=9x+4\)
=> \(\left\{{}\begin{matrix}x\ge-\frac{4}{9}\\4\left(4x^2+5x-1\right)=81x^2+72x+16\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ge-\frac{4}{9}\\65x^2+52x+20=0\end{matrix}\right.\)
=> PT vô nghiệm
Vậy PT vô nghiệm
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
cái = 0 của pt 2 ý,,,,bạn thấy nha,,,do x>0 ( ĐKXĐ) ta có \(\frac{5\left(x+49\right)}{\sqrt{5x^2+4x}+21}\ge\frac{x+6}{\sqrt{x^2-3x-18}+6}\)
Từ đó dẫn đến vô lí
b)\(\sqrt{5x^2+4x}-\sqrt{x^2-3x-18}=5\sqrt{x}\)
Đk:....
\(\Leftrightarrow\sqrt{5x^2+4x}-21-\left(\sqrt{x^2-3x-18}-6\right)-\left(5\sqrt{x}-15\right)=0\)
\(\Leftrightarrow\frac{5x^2+4x-441}{\sqrt{5x^2+4}+21}-\frac{x^2-3x-18-36}{\sqrt{x^2-3x-18}+6}-\frac{25x-225}{5\sqrt{x}+15}=0\)
\(\Leftrightarrow\frac{\left(x-9\right)\left(5x+49\right)}{\sqrt{5x^2+4}+21}-\frac{\left(x-9\right)\left(x+6\right)}{\sqrt{x^2-3x-18}+6}-\frac{25\left(x-9\right)}{5\sqrt{x}+15}=0\)
\(\Leftrightarrow\left(x-9\right)\left(\frac{5x+49}{\sqrt{5x^2+4}+21}-\frac{x+6}{\sqrt{x^2-3x-18}+6}-\frac{25}{5\sqrt{x}+15}\right)=0\)
chịu cái trong ngoặc r` bình phương đi :v
\(\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=9x-3\)
\(\Leftrightarrow\sqrt{4x^2+5x+1}-\dfrac{2\sqrt{7}}{3}-\left(2\sqrt{x^2-x+1}-\dfrac{2\sqrt{7}}{3}\right)=9x-3\)
\(\Leftrightarrow\dfrac{4x^2+5x+1-\dfrac{28}{9}}{\sqrt{4x^2+5x+1}+\dfrac{2\sqrt{7}}{3}}-\dfrac{4\left(x^2-x+1\right)-\dfrac{28}{9}}{2\sqrt{x^2-x+1}+\dfrac{2\sqrt{7}}{3}}=9x-3\)
\(\Leftrightarrow\dfrac{\dfrac{36x^2+45x-19}{9}}{\sqrt{4x^2+5x+1}+\dfrac{2\sqrt{7}}{3}}-\dfrac{\dfrac{36x^2-36x+8}{9}}{2\sqrt{x^2-x+1}+\dfrac{2\sqrt{7}}{3}}=3\left(3x-1\right)\)
\(\Leftrightarrow\dfrac{\dfrac{\left(3x-1\right)\left(12x+19\right)}{9}}{\sqrt{4x^2+5x+1}+\dfrac{2\sqrt{7}}{3}}-\dfrac{\dfrac{4\left(3x-2\right)\left(3x-1\right)}{9}}{2\sqrt{x^2-x+1}+\dfrac{2\sqrt{7}}{3}}-3\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(\dfrac{12x+19}{9\left(\sqrt{4x^2+5x+1}+\dfrac{2\sqrt{7}}{3}\right)}-\dfrac{4\left(3x-2\right)}{9\left(2\sqrt{x^2-x+1}+\dfrac{2\sqrt{7}}{3}\right)}-3\right)=0\)
Dễ thấy: \(\dfrac{12x+19}{9\left(\sqrt{4x^2+5x+1}+\dfrac{2\sqrt{7}}{3}\right)}-\dfrac{4\left(3x-2\right)}{9\left(2\sqrt{x^2-x+1}+\dfrac{2\sqrt{7}}{3}\right)}-3< 0\)
\(\Rightarrow3x-1=0\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)
Ace Legona cái dễ thấy của bạn mình nghĩ lại là mấu chốt của bài này
\(\sqrt{4x^2+5x+1}-2\sqrt{x^2-x+1}=9x-3\)
Đặt \(\hept{\begin{cases}\sqrt{4x^2+x+1}=a\\\sqrt{x^2-x+1}=b\end{cases}}\) \(\left(a,b\ge00\right)\)
Khi đó có pt \(a-2b=a^2-4b^2\)
\(\Leftrightarrow-\left(a-2b\right)\left(a+2b-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}b=\frac{1}{2}-\frac{a}{2}\\b=\frac{a}{2}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}\sqrt{x^2-x+1}=\frac{1}{2}-\frac{\sqrt{4x^2+x+1}}{2}\\\sqrt{x^2-x+1}=\frac{\sqrt{4x^2+x+1}}{2}\end{cases}}\)\(\Rightarrow x=\frac{1}{3}\)
Bạn tham khảo:
Câu hỏi của Nguyễn Thị Bình Yên - Toán lớp 9 | Học trực tuyến
Bạn lưu ý:
\(a=\sqrt{4x^2+5x+1}\ge0\)
\(b=\sqrt{4x^2-4x+4}=\sqrt{\left(2x-1\right)^2+3}\ge\sqrt{3}>1\)
Do đó \(a+b>1\) hay \(a+b-1>0\)