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\(d,x-5\sqrt{x}=0\)
\(ĐKXĐ:x\ge0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\\sqrt{x}-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\sqrt{x}=5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}\)(Thỏa mãn ĐKXĐ)
Vậy...
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a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)
b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)
c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)
\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)
d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)
\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)
a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)
<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)
<=> \(\sqrt{x}+8=28\)
<=> \(\sqrt{x}=28-8\)
<=> \(\sqrt{x}=20\)
<=> \(\left(\sqrt{x}\right)^2=20^2\)
<=> x = 400
=> x = 400
b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)
<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)
<=> \(3\sqrt{x}+5=\sqrt{x}+12\)
<=> \(3\sqrt{x}=\sqrt{x}+12-5\)
<=> \(3\sqrt{x}=\sqrt{x}+7\)
<=> \(3\sqrt{x}-\sqrt{x}=7\)
<=> \(2\sqrt{x}=7\)
<=> \(\sqrt{x}=\frac{7}{2}\)
<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)
<=> \(x=\frac{49}{4}\)
=> \(x=\frac{49}{4}\)
c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)
<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)
<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)
<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)
<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)
<=> \(8\sqrt{x}=6\sqrt{x}+4\)
<=> \(8\sqrt{x}-6\sqrt{x}=4\)
<=> \(2\sqrt{x}=4\)
<=> \(\sqrt{x}=2\)
<=> \(\left(\sqrt{x}\right)^2=2^2\)
<=> x = 4
=> x = 4
d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)
<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)
<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)
<=>\(2\sqrt{3x}=6\sqrt{3x}\)
<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)
<=>\(-4\sqrt{3x}=0\)
<=> \(\sqrt{3x}=0\)
<=> \(\left(\sqrt{3x}\right)^2=0^2\)
<=> 3x = 0
<=> x = 0
=> x = 0
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a và b chắc của lớp 9 nhỉ
\(x^2-2x+2=x^2-x-x+2\)
\(=x\left(x-1\right)-\left(x-1\right)+1\)
\(=\left(x-1\right)^2+1\)
\(9x^2-6x+5=9\left(x^2-\frac{2}{3}x+\frac{5}{9}\right)\)
\(=9\left(x^2-\frac{1}{3}x-\frac{1}{3}x+\frac{5}{9}\right)\)
\(=9\left(x^2-\frac{1}{3}x-\frac{1}{3}x+\frac{1}{9}+\frac{4}{9}\right)\)
\(=9\left[x\left(x-\frac{1}{3}\right)-\frac{1}{3}\left(x-\frac{1}{3}\right)+\frac{4}{9}\right]\)
\(=9\left[\left(x-\frac{1}{3}\right)^2+\frac{4}{9}\right]\)
\(=9\left(x-\frac{1}{3}\right)^2+4\)
Cái kia tương tự.
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Dựa vào đây mà làm nhé : Câu hỏi của nhi anny - Toán lớp 9 - Học toán với OnlineMath
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Làm hơi tắt , thông cảm ;))
Từ (1) \(\Rightarrow36=\left(x+y+z\right)^2\Leftrightarrow36=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)
\(\Leftrightarrow36=18+2\left(xy+yz+zx\right)\Leftrightarrow xy+yz+zx=9\)(4)
Từ (3) \(\Rightarrow16=\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2\Leftrightarrow16=x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)\)
\(\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=5\Leftrightarrow\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2=25\)
\(\Leftrightarrow xy+yz+zx+2\left(\sqrt{xy^2z}+\sqrt{xyz^2}+\sqrt{x^2yz}\right)=25\)
\(\Leftrightarrow\sqrt{xyz}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)=8\Leftrightarrow\sqrt{xyz}=\frac{8}{4}\Leftrightarrow xyz=4\)(5)
Vậy hệ đã cho tương đương với :
\(\hept{\begin{cases}x+y+z=6\left(1\right)\\xy+yz+zx=9\left(4\right)\\xyz=4\left(5\right)\end{cases}}\)
Từ (5) \(\Rightarrow yz=\frac{4}{x}\)(Dễ thấy \(x,y,z>0\))
(4) \(\Leftrightarrow xy+yz+zx+x^2=9+x^2\Leftrightarrow x\left(x+y+z\right)+yz=9+x^2\)
\(\Leftrightarrow x.6+\frac{4}{x}=9+x^2\Leftrightarrow x^3-6x^2+9x-4=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x-4\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}.}\)
Thế vào ta suy ra hệ có các nghiệm : \(\left(x,y,z\right)=\left(1,1,4\right),\left(1,4,1\right),\left(4,1,1\right).\)