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Dk: x\(\ge0\)
lien hop
\(\Leftrightarrow\sqrt{x+3}-\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{x+3}=2\Rightarrow x=1\)
Điều kiện xác định
\(\hept{\begin{cases}2-x^2+2x\ge0\\-x^2-6x-8\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}-0,73\le x\le2,73\\-4\le x\le-2\end{cases}}\)
=> Tập xác định là tập rỗng
Vậy pt vô nghiệm
2, \(\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9x+9}+24\sqrt{\dfrac{x+1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{9}{2}\sqrt{x+1}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow-\sqrt{x+1}=-17\)
\(\Leftrightarrow x+1=289\left(x>0\right)\)
\(\Leftrightarrow x=288\)
Vậy x = 288
3, \(-5x+7\sqrt{x}+12=0\)
\(\Leftrightarrow-5x+12\sqrt{x}-5\sqrt{x}+12=0\)
\(\Leftrightarrow\sqrt{x}\left(12-5\sqrt{x}\right)+\left(12-5\sqrt{x}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(12-5\sqrt{x}\right)=0\)
Do \(\sqrt{x}+1>0\)
\(\Rightarrow12-5\sqrt{x}=0\Leftrightarrow x=\dfrac{144}{25}\)
Vậy...
1. (Đề có chút sai sai nên mình sửa lại nhé) \(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
(ĐK: \(x\ge1\))
\(\Leftrightarrow\sqrt{36\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{4\left(x-1\right)}=16-\sqrt{x-1}\)
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}+\sqrt{x-1}=16\)
\(\Leftrightarrow2\sqrt{x-1}=16\)
\(\Leftrightarrow\sqrt{x-1}=8\)
\(\Leftrightarrow x-1=64\)
\(\Leftrightarrow x=65\left(tm\right)\)
Vậy pt đã cho có nghiệm x=65.
2. \(\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9x+9}+24\sqrt{\dfrac{x+1}{64}}=-17\)
(ĐK: \(x\ge-1\))
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{3}{2}\sqrt{9\left(x+1\right)}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x+1}-\dfrac{9}{2}\sqrt{x+1}+3\sqrt{x+1}=-17\)
\(\Leftrightarrow-\sqrt{x+1}=-17\)
\(\Leftrightarrow\sqrt{x+1}=17\)
\(\Leftrightarrow x+1=289\)
\(\Leftrightarrow x=288\left(tm\right)\)
Vậy \(S=\left\{288\right\}\)
3. \(-5x+7\sqrt{x}+12=0\) (ĐK: \(x\ge0\))
\(\Leftrightarrow5x-7\sqrt{x}-12=0\)
\(\Leftrightarrow5x+5\sqrt{x}-12\sqrt{x}-12=0\)
\(\Leftrightarrow5\sqrt{x}\left(\sqrt{x}+1\right)-12\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(5\sqrt{x}-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\5\sqrt{x}-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(vô.lý\right)\\5\sqrt{x}=12\end{matrix}\right.\Leftrightarrow\sqrt{x}=\dfrac{12}{5}\Leftrightarrow x=\dfrac{144}{25}\left(tm\right)\)
Vậy pt có nghiệm \(x=\dfrac{144}{25}\)
a) \(x^2+3-\sqrt{2x^2-3x+2}=\frac{3}{2}\left(x+1\right)\)
\(\Leftrightarrow x^2.2+3.2-\sqrt{2x^2-3x+2}.3=\frac{3}{2}\left(x+1\right).2\)
\(\Leftrightarrow2x^2+6-\sqrt{2x^2-3x+2}=3\left(x+1\right)\)
\(\Leftrightarrow2x^2+6-2\sqrt{2x^2-3x+2}=3x+3\)
\(\Leftrightarrow-2\sqrt{2x^2-3x+2}+6=3x^2+3-2x^2\)
\(\Leftrightarrow-2\sqrt{2x^2-3x+2}=3x+3-2x^2-6\)
\(\Leftrightarrow-2\sqrt{2x^2-3x+2}=-2x^3+3x-3\)
\(\Leftrightarrow\left(-2\sqrt{2x^2-3x+2}\right)^2=\left(-2x^2+3x-3\right)^2\)
\(\Leftrightarrow8x^2-12x+8=4x^4-12x^3+21x^2-18x+9\)
\(\Leftrightarrow4x^2-12x^3+12x^2-6x+1=0\)
\(\Leftrightarrow\left(x-2\right)^2\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: nghiệm phương trình là \(\left\{1;\frac{1}{2}\right\}\)
b) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}-2\right)^2}+\sqrt{\left(\sqrt{x-1}-3\right)^2}=1\)
\(\Leftrightarrow\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|=1\)
Xét \(\left|\sqrt{x-1}-2\right|+\left|\sqrt{x-1}-3\right|\)
\(=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=\left|1\right|=1\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x-1}-2\right)\left(3-\sqrt{x-1}\right)\ge0\Leftrightarrow5\le x\le10\)
\(\sqrt{x+8}=\sqrt{3x+2}+\sqrt{x+3}\) dkxd \(\left\{{}\begin{matrix}x\ge-8\\x\ge\\x\ge-\dfrac{2}{3}\end{matrix}\right.-3\)=>x\(\ge\)\(\dfrac{-2}{3}\)
\(x+8=3x+2+x+3+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(x+8=4x+5+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(x+8-4x-5=2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
-3x+3=\(2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(\left\{{}\begin{matrix}-3\left(x-3\right)\ge0\\\left(-3x+3\right)^2=4.\left(3x+2\right)\left(x+3\right)\end{matrix}\right.\)
Chắc tới đây bạn làm đc rồi nhỉ
6.
Đặt \(\left\{{}\begin{matrix}\sqrt{5x^2+6x+5}=a\\4x=b\end{matrix}\right.\)
\(\Rightarrow a\left(a^2+1\right)=b\left(b^2+1\right)\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{5x^2+6x+5}=4x\left(x\ge0\right)\)
\(\Leftrightarrow5x^2+6x+5=16x^2\)
\(\Leftrightarrow11x^2-6x-5=0\)
\(\Rightarrow x=1\)
4. Bạn coi lại đề (chính xác là pt này ko có nghiệm thực)
5.
\(\Leftrightarrow x^2+x+6-\left(2x+1\right)\sqrt{x^2+x+6}+6x-6=0\)
Đặt \(\sqrt{x^2+x+6}=t>0\)
\(t^2-\left(2x+1\right)t+6x-6=0\)
\(\Delta=\left(2x+1\right)^2-4\left(6x-6\right)=\left(2x-5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{2x+1+2x-5}{2}=2x-2\\t=\frac{2x+1-2x+5}{2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+6}=2x-2\left(x\ge1\right)\\\sqrt{x^2+x+6}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=4x^2-8x+4\left(x\ge1\right)\\x^2+x+6=9\end{matrix}\right.\)
Câu hỏi của tran huu dinh - Toán lớp 8 - Học toán với OnlineMath
Đây nè bạn
mơn bạn mik cũng đặt ẩn phụ hoàn toàn
zậy bạn lm giúp mik hai câu cúi nhé!!!!
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow2x^2+5x-1=7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow2\left(x^2+x+1\right)+3\left(x-1\right)-7\sqrt{\left(x-1\right)\left(x^2+x+1\right)}=0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+x+1}=a\\\sqrt{x-1}=b\end{matrix}\right.\)
\(\Rightarrow2a^2+3b^2-7ab=0\)
\(\Leftrightarrow\left(a-3b\right)\left(2a-b\right)=0\Leftrightarrow\left[{}\begin{matrix}a=3b\\2a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+1}=3\sqrt{x-1}\\2\sqrt{x^2+x+1}=\sqrt{x-1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=9\left(x-1\right)\\4\left(x^2+x+1\right)=x-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
b/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\sqrt{x-2}-\sqrt{x+2}+2x-2\sqrt{x^2-4}-2=0\)
Đặt \(\sqrt{x-2}-\sqrt{x+2}=a< 0\)
\(\Rightarrow a^2=2x-2\sqrt{x^2-4}\)
Phương trình trở thành:
\(a+a^2-2=0\Leftrightarrow\left[{}\begin{matrix}a=1\left(l\right)\\a=-2\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x-2}-\sqrt{x+2}=-2\)
\(\Leftrightarrow\sqrt{x-2}+2=\sqrt{x+2}\)
\(\Leftrightarrow x+2+4\sqrt{x-2}=x+2\)
\(\Leftrightarrow\sqrt{x-2}=0\)