giai phuong trinh \(\sqrt{2x+4-6\sqrt{2x-5}}+\sqrt{2x-4+2\sqrt{2x-5}}=4\)
Những câu hỏi liên quan
Giai phuong trinh
\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-2\sqrt{2x-5}}=2\sqrt{2}\)
\(\sqrt{x+2+3\sqrt{ }2x-5}+\sqrt{x-2-\sqrt{ }2x-5}=2\sqrt{2}\) 2\(\sqrt{2}\) giai phuong trinh
giai cac phuong trinh
a)\(2x^4+5x^3+x^2+5x+2=0\)
b)\(\sqrt{x-1}-\sqrt[3]{2-x}=1\)
c)\(x-\sqrt{x}+1=\sqrt{2x^2-30x+2}\)
d)\(2x^2+3x+7=\left(x-5\right)\sqrt{2x^2+1}\)
e)\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
giai phuong trinh:
\(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)
ĐKXĐ : \(-4\le x\le4\)
TA CÓ : \(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)
\(\Leftrightarrow\left[\left(\sqrt{x+4}-2\right)\left(\sqrt{x+4}+2\right)\right]\left(\sqrt{4-x}+2\right)=2x\left(\sqrt{x+4}+2\right)\)
\(\Leftrightarrow\left[x+4-4\right]\left(\sqrt{4-x}+2\right)-2x\left(\sqrt{x+4}+2\right)=0\)
\(\Leftrightarrow x\left(\sqrt{4-x}+2\right)-2x\left(\sqrt{x+4}+2\right)=0\)
\(\Leftrightarrow x\left[\sqrt{4-x}+2-2\sqrt{x+4}-4\right]=0\)
\(\Leftrightarrow x=0\)HOẶC \(\sqrt{4-x}-2\sqrt{x+4}-2=0\)
VỚI \(\sqrt{4-x}-2\sqrt{x+4}-2=0\)
\(\Leftrightarrow\sqrt{4-x}-2=2\sqrt{x+4}\)
\(\Leftrightarrow4-x+4-4\sqrt{4-x}=4x+16\)
\(\Leftrightarrow8-x-4x-16=4\sqrt{4-x}\)
\(\Leftrightarrow-5x-8=4\sqrt{4-x}\)ĐK : \(-4\le x\le\frac{-8}{5}\)
\(\Leftrightarrow\left[-\left(5x+8\right)\right]^2=16\left(4-x\right)\)
\(\Leftrightarrow25x^2+64+80x=64-16x\)
\(\Leftrightarrow25x^2+96x=0\Leftrightarrow x\left(25x+96\right)=0\)
\(\Leftrightarrow x=0\)HOẶC \(x=\frac{-96}{25}\)(THỎA MÃN ĐK )
VẬY PT CÓ 2 NGHIỆM \(x\in\left[0;\frac{-96}{25}\right]\)
P/S : CÁCH CỦA MÌNH KHÁ DÀI VÀ CHI TIẾT QUÁ . BẠN CÓ THỂ THAM KHẢO CÁCH KHÁC NHANH HƠN :>
Đúng 0
Bình luận (0)
Giai phuong trinh \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
Giai phuong trinh
1/ \(\sqrt{x-3}+\sqrt{2-x}=5\)
2/ \(2x+7\sqrt{x}+\dfrac{7}{\sqrt{x}}+\dfrac{2}{x}+9=0\)
3/ \(x+\dfrac{1}{x}-4\sqrt{x}-\dfrac{4}{\sqrt{x}}+6=0\)
4/ \(\sqrt{x+9}=5-\sqrt{x-2}\)
giai phuong trinh: \(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x-1}\)
giai phuong trinh \(x^2-5x+4=2\sqrt{2x-4}\)
\(ĐK:x\ge2\)
\(x^2-5x+4=2\sqrt{2x-4}\)
<=>\(x^2-5x+4=2\sqrt{2\left(x-2\right)}\)
<=>\(x^2-5x+4+x-2+2=\left(x-2\right)+2\sqrt{2\left(x-2\right)}+2\)
<=>\(x^2-4x+4=\left(\sqrt{x-2}+2\right)^2\)
<=>\(\left(x-2\right)^2=\left(\sqrt{x-2}+2\right)^2\)
<=> \(\left(x-2-\sqrt{x-2}-2\right)\left(x-2+\sqrt{x-2}+2\right)=0\)
<=>\(\left(x-\sqrt{x-2}-4\right)\left(x+\sqrt{x-2}\right)=0\)
Xét \(x-\sqrt{x-2}-4=0\)
<=>\(x^2-8x+16=x-2\)
<=>\(x^2-9x+18=0\)
=> x=6;3(nhận)
Xet1\(x+\sqrt{x-2}=0\)
Do x\(\ge2\)=> pt vô nghiệm
Vậy ...
Đúng 0
Bình luận (0)
giai phuong trinh : \(2x^2\left(5-\sqrt[3]{5x-x^3}\right)=2x^3+17x-8\)