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a, \(5\sqrt{2x^2+3x+9}=2x^2+3x+3\) (*)
Đặt \(2x^2+3x=a\left(a\ge-9\right)\)
=> \(5\sqrt{a+9}=a+3\)
<=> \(25\left(a+9\right)=a^2+6a+9\)
<=> \(25a+225=a^2+6a+9\)
<=> \(0=a^2+6a+9-25a-225=a^2-19a-216\)
<=> 0= \(a^2-27a+8a-216\)
<=> \(\left(a-27\right)\left(a+8\right)=0\)
=> \(\left[{}\begin{matrix}a=27\\a=-8\end{matrix}\right.\) <=>\(\left[{}\begin{matrix}2x^2+3x=27\\2x^2+3x=-8\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}2x^2+3x-27=0\\2x^2+3x+8=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\left(x-3\right)\left(2x+9\right)=0\\2\left(x^2+2.\frac{3}{4}+\frac{9}{16}\right)+\frac{55}{8}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=3\left(tm\right)\\x=-\frac{9}{2}\left(tm\right)\\2\left(x+\frac{3}{4}\right)^2=-\frac{55}{8}\left(ktm\right)\end{matrix}\right.\)
Vậy pt (*) có tập nghiệm \(S=\left\{3,-\frac{9}{2}\right\}\)
b, \(9-\sqrt{81-7x^3}=\frac{x^3}{2}\left(đk:x\le\sqrt[3]{\frac{81}{7}}\right)\)(*)
<=> \(\sqrt{81-7x^3}=9-\frac{x^3}{2}\)
<=>\(81-7x^3=\left(9-\frac{x^3}{2}\right)^2=81-9x^3+\frac{x^6}{4}\)
<=> \(-7x^3+9x^3-\frac{x^6}{4}=0\) <=> \(2x^3-\frac{x^6}{4}=0\)<=> \(8x^3-x^6=0\)
<=> \(x^3\left(8-x^2\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\8=x^2\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=\pm2\sqrt{2}\left(ktm\right)\end{matrix}\right.\)
Vậy pt (*) có nghiệm x=0
d,\(\sqrt{9x-2x^2}-9x+2x^2+6=0\) (*) (đk: \(0\le x\le\frac{1}{2}\))
<=> \(\sqrt{9x-2x^2}-\left(9x-2x^2\right)+6=0\)
Đặt \(\sqrt{9x-2x^2}=a\left(a\ge0\right)\)
Có \(a-a^2+6=0\)
<=> \(a^2-a-6=0\) <=> \(a^2-3x+2x-6=0\)
<=> \(\left(a-3\right)\left(a+2\right)=0\)
=> \(a-3=0\) (vì a+2>0 vs mọi \(a\ge0\))
<=> a=3 <=>\(\sqrt{9x-2x^2}=3\) <=> \(9x-2x^2=9\)
<=> 0=\(2x^2-9x+9\) <=> \(2x^2-6x-3x+9=0\) <=>\(\left(2x-3\right)\left(x-3\right)=0\)
=> \(\left[{}\begin{matrix}2x=3\\x=3\end{matrix}\right.< =>\left[{}\begin{matrix}x=\frac{3}{2}\\x=3\end{matrix}\right.\)(t/m)
Vậy pt (*) có tập nghiệm \(S=\left\{\frac{3}{2},3\right\}\)
a/ Giải rồi
b/ ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)
Pt trở thành:
\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)
\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)
\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)
\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)
\(\Leftrightarrow...\)
e/ ĐKXD: \(x>0\)
\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)
\(\Rightarrow t^2=x+\frac{1}{4x}+1\)
Pt trở thành:
\(5t=2\left(t^2-1\right)+4\)
\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)
\(\Leftrightarrow2x-4\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)
\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)
(1)Phương trình đã cho tương đương với:
√3x2−7x+3−√3x2−5x−1=√x2−2−√x2−3x+43x2−7x+3−3x2−5x−1=x2−2−x2−3x+4
⇔−2x+4√3x2−7x+3+√3x2−5x−1=3x−6√x2−2+√x2−3x
a) \(\text{Đ}K\text{X}\text{Đ}:\frac{3}{2}\le x\le\frac{5}{2}\)
Áp dụng BĐT Bunhiacopxki ta có:
\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)
Dấu '=' xảy ra khi \(\sqrt{2x-3}=\sqrt{5-2x}\Leftrightarrow x=2\)
Lại có: \(VP=3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)
Dấu '=' xảy ra khi x=2
Do đó VT=VP khi x=2
b) ĐK: \(x\ge0\). Ta thấy x=0 k pk là nghiệm của pt, chia 2 vế cho x ta có:
\(x^2-2x-x\sqrt{x}-2\sqrt{x}+4=0\Leftrightarrow x-2-\sqrt{x}-\frac{2}{\sqrt{x}}+\frac{4}{x}=0\)
\(\Leftrightarrow\left(x+\frac{4}{x}\right)-\left(\sqrt{x}+\frac{2}{\sqrt{x}}\right)-2=0\)
Đặt \(\sqrt{x}+\frac{2}{\sqrt{x}}=t>0\Leftrightarrow t^2=x+4+\frac{4}{x}\Leftrightarrow x+\frac{4}{x}=t^2-4\), thay vào ta có:
\(\left(t^2-4\right)-t-2=0\Leftrightarrow t^2-t-6=0\Leftrightarrow\left(t-3\right)\left(t+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t=3\\t=-2\end{cases}}\)
Đối chiếu ĐK của t
\(\Rightarrow t=3\Leftrightarrow\sqrt{x}+\frac{2}{\sqrt{x}}=3\Leftrightarrow x-3\sqrt{x}+2=0\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=1\end{cases}}\)
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
bạn giải theo delta nha :) mình vd một câu đó
\(1.x^2-11x+30=0\)
\(\Delta=\left(-11\right)^2-4.1.30=1>0\)
Do đó pt có 2 nghiệm phân biệt là:
\(x_1=\frac{11+\sqrt{1}}{2}=6;x_2=\frac{11-\sqrt{1}}{2}=5\)
đkxđ: x khác -3; x khác 3
\(\frac{x^2-x}{x+3}-\frac{x^2}{x-3}=\frac{7x^2-3x}{9-x^2}\)
\(\frac{\left(x^2-x\right)\left(x-3\right)-x^2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=-\frac{7x^2-3x}{x^2-9}\)
\(\frac{x^3-3x^2-x^2+3x-x^3-3x^2}{\left(x-3\right)\left(x+3\right)}=-\frac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(\frac{-7x^2+3x}{\left(x-3\right)\left(x+3\right)}=-\frac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(-\frac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}-\frac{-7x^2+3x}{\left(x-3\right)\left(x+3\right)}=0\)
\(\frac{-7x^2+3x+7x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\)
\(\frac{0}{\left(x-3\right)\left(x+3\right)}=0\)
kết luận: \(x\in R\ne\pm3\)
Lsao vậy