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Câu 1 là \(\left(8x-4\right)\sqrt{x}-1\) hay là \(\left(8x-4\right)\sqrt{x-1}\)?
Câu 1:ĐK \(x\ge\frac{1}{2}\)
\(4x^2+\left(8x-4\right)\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
<=> \(\left(4x^2-3x-1\right)+4\left(2x-1\right)\sqrt{x}-2\sqrt{\left(2x-1\right)\left(x+3\right)}\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}\left(2\sqrt{x\left(2x-1\right)}-\sqrt{x+3}\right)=0\)
<=> \(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{8x^2-4x-x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=>\(\left(x-1\right)\left(4x+1\right)+2\sqrt{2x-1}.\frac{\left(x-1\right)\left(8x+3\right)}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}=0\)
<=> \(\left(x-1\right)\left(4x+1+2\sqrt{2x-1}.\frac{8x+3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}\right)=0\)
Với \(x\ge\frac{1}{2}\)thì \(4x+1+2\sqrt{2x-1}.\frac{8x-3}{2\sqrt{x\left(2x-1\right)}+\sqrt{x+3}}>0\)
=> \(x=1\)(TM ĐKXĐ)
Vậy x=1
Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
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. \(9x=225\Rightarrow x=25\)
b. \(2x=8\Rightarrow x=4\)
c. \(3\left(2x-3\right)=6\Rightarrow6x=15\Rightarrow x=\frac{15}{6}\)
d. \(4\left(x+1\right)=8\Rightarrow4x=4\Rightarrow x=1\)
e. \(\sqrt{x+2}.\sqrt{x-2}-\sqrt{x-2}=0\Rightarrow\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
=> \(\sqrt{x-2}=0\Rightarrow x-2=0\Rightarrow x=2\)
hoặc \(\sqrt{x+2}-1=0\Rightarrow\sqrt{x+2}=1\Rightarrow x+2=1\Rightarrow x=-1\)
f. \(\sqrt{x+1}+3\sqrt{x+1}=4\Rightarrow4\sqrt{x+1}=4\Rightarrow\sqrt{x+1}=1\Rightarrow x+1=1\Rightarrow x=0\)
g. \(\sqrt{x-2}\left(1-\sqrt{x}\right)=0\)
=> \(\sqrt{x-2}=0\Rightarrow x-2=0\Rightarrow x=2\)
hoặc \(1-\sqrt{x}=0\Rightarrow\sqrt{x}=1\Rightarrow x=1\)
h. \(\sqrt{x+3}.\sqrt{x-3}-\sqrt{x+3}=0\Rightarrow\sqrt{x+3}\left(\sqrt{x-3}-1\right)=0\)
=> \(\sqrt{x+3}=0\Rightarrow x=-3\)
hoặc \(\sqrt{x-3}-1=0\Rightarrow\sqrt{x-3}=1\Rightarrow x=4\)
√x−2(1−√x)=0
=> √x−2=0⇒x−2=0⇒x=2
hoặc 1−√x=0⇒√x=1⇒x=1
h. √x+3.√x−3−√x+3=0⇒√x+3(√x−3−1)=0
=> √x+3=0⇒x=−3
hoặc
√x−2(1−√x)=0
=> √x−2=0⇒x−2=0⇒x=2
hoặc 1−√x=0⇒√x=1⇒x=1
h. √x+3.√x−3−√x+3=0⇒√x+3(√x−3−1)=0
=> √x+3=0⇒x=−3
hoặc
nhờ casio và 1 số suy đoán ta biết được max f(x) =7 khi x=0 ,giờ AM-GM ngược thôi :v
ta có: \(f\left(x\right)=\sqrt{\left(2x+3\right)\left(x+3\right)}+\sqrt{4\left(x+4\right)}-2x\)
Áp dụng bất đẳng thức cauchy :
\(\sqrt{\left(2x+3\right)\left(x+3\right)}\le\frac{1}{2}\left(3x+6\right)\)
\(\sqrt{4\left(x+4\right)}\le\frac{1}{2}\left(x+8\right)\)
\(\Rightarrow f\left(x\right)\le\frac{1}{2}\left(4x+14\right)-2x=2x+7-2x=7\)
đẳng thức xảy ra khi \(\hept{\begin{cases}2x+3=x+3\\4=x+4\end{cases}\Leftrightarrow x=0}\)
Còn ý liền trước nó nữa:
Tìm tất cả các cặp số (x, y) thỏa mãn \(2\left(x\sqrt{y-4}+y\sqrt{x-4}\right)=xy\)
LÀM GIÚP MK CÂU TÌM GTLN NHA
HELP ME, PLEASE!