a) Giai phuong trinh sau: \(\sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0\)
b) Tim so tu nhien x, y thoa man: x( 1 + x +x2) = 4y( y - 1)
a)Tim tat ca cac so nguyen duong x, y , z thoa man: \(\frac{x+y\sqrt{2013}}{y+z\sqrt{2013}}\)la so huu ti, dong thoi x2 + y2+ z2 la so nguyen to.
b) Tim so tu nhien x, y thoa man: x(1+x+x2) = y(y-1).
Giai phuong trinh sau: \(\sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0\)
\(\sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0;ĐK:x\ge4\)
\(\Leftrightarrow\sqrt{x}+\sqrt{x+9}=\sqrt{x+1}-\sqrt{x+4}\)
\(\Leftrightarrow2x+9+2\sqrt{x^2+9x}=2x-5+2\sqrt{x^2-5x+4}\)
\(\leftrightarrow14+2\sqrt{x^2+9x}=2\sqrt{x^2-5x+4}\leftrightarrow7+\sqrt{x^2+9x}=\sqrt{x^2-5x+4}\)
\(\leftrightarrow49+14\sqrt{x^2+9x}+x^2+9x=x^2-5x+4\)
\(\leftrightarrow14\sqrt{x^2+9x}=-14x-45\)
\(\leftrightarrow\hept{\begin{cases}196.x^2+9x=196x^2+1260x+2025\\-14x-45\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}504x=2025\\x\le\frac{-45}{14}\end{cases}\leftrightarrow x=\frac{225}{56}}\) loại
-> PT vô nghiệm
Giai phuong trinh sau: \(\sqrt{x}-\sqrt{x+1}-\sqrt{x+4}+\sqrt{x+9}=0\)
bai1 giai phuong trinh 2\(\sqrt{x}\)- \(\sqrt{3x+1}\)= x-1
bai2 cho 2 so thuc duong A , B thoa man A+B+3AB=1.
tim MAX P=\(\sqrt{1-A^2}\)+\(\sqrt{1-B^2}\)+\(\frac{3AB}{A+B}\)
Giai phuong trinh va he phuong trinh:
a) \(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)
b) \(x^2+3x+1=\left(x+3\right).\sqrt{x^2+1}\)
c) \(\left\{{}\begin{matrix}x^2+y^2=11\\x+xy+y=3+4\sqrt{2}\end{matrix}\right.\)
cho phuong trinh:\(\dfrac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\dfrac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
a/tim dieu kien cua x de phuong trinh co nghia
b/giai phuong trinh
a: ĐKXĐ: x>=0
b: \(\Leftrightarrow\dfrac{2\sqrt{2}-2\sqrt{2-\sqrt{x}}+\sqrt{2x}-\sqrt{x\left(2-\sqrt{x}\right)}+2\sqrt{2}+2\sqrt{2+\sqrt{x}}-\sqrt{2x}-\sqrt{x\left(2+\sqrt{x}\right)}}{2-2+\sqrt{x}}=\sqrt{2}\)
\(\Leftrightarrow4\sqrt{2}-2\sqrt{x\left(\sqrt{x}+2\right)}=\sqrt{2x}\)
\(\Leftrightarrow\sqrt{4x\left(\sqrt{x}+2\right)}=4\sqrt{2}-\sqrt{2x}\)
\(\Leftrightarrow4x\left(\sqrt{x}+2\right)=32-16\sqrt{x}+2x\)
\(\Leftrightarrow4x\sqrt{x}+8x-32+16\sqrt{x}-2x=0\)
=>\(x\in\left\{0;1.2996\right\}\)
Giai he phuong trinh:
a) \(\left\{{}\begin{matrix}x^2-y^2=1\\4x^2-5xy=2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x+\sqrt{y+2018}=1\\\sqrt{x+2018}+y=1\end{matrix}\right.\)
c) \(\left\{{}\begin{matrix}x+y=\sqrt{4z-1}\\y+z=\sqrt{4x-1}\\z+x=\sqrt{4y-1}\end{matrix}\right.\)
giai phuong trinh sau:
\(\sqrt{x+3+4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=5\)
Áp dụng BĐT:\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
Ta có: \(\left|\sqrt{x-1}+2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}+2+3-\sqrt{x-1}\right|=5\)
Dấu \(=\)xảy ra khi \(AB\ge0\)
dat \(\sqrt{x-1}\) = t
ta có: \(\sqrt{x+3+4t}\)+ \(\sqrt{x+8-6t}\)= 5
x + 3 + 4t + x + 8 - 6t = 25
2x - 2t = 14 ( chia cả 2 vế cho 2)
x - t = 7
t = x - 7
thay t = \(\sqrt{x}-1\)vào ta được:
x - 7 = \(\sqrt{x-1}\)
( x - 7 )2 = x - 1
x2 -14x + 49 = x - 1
x2 - 15x + 50 = 0
k biết đúng hay k
OoO Ledegill2 OoO. Ban co the giai thich ro hon giup minh duoc khong. hi
Giai phuong trinh: \(\sqrt{3x+x^2+\dfrac{9}{4}}+\sqrt{x^2+3x+1}=0\)
Lời giải:
Với mọi $x$ thuộc ĐKXĐ, ta luôn có:
\(\left\{\begin{matrix} \sqrt{3x+x^2+\frac{9}{4}}\geq 0\\ \sqrt{x^2+3x+1}\geq 0\end{matrix}\right.\)
Do đó, để \(\sqrt{3x+x^2+\frac{9}{4}}+\sqrt{x^2+3x+1}=0\) thì:
\(\left\{\begin{matrix} \sqrt{3x+x^2+\frac{9}{4}}= 0\\ \sqrt{x^2+3x+1}=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x=\frac{-3}{2}\\ x=\frac{3\pm \sqrt{5}}{2}\end{matrix}\right.\) (vô lý)
Do đó pt vô nghiệm.