giai phuong trinh : \(2x^2\left(5-\sqrt[3]{5x-x^3}\right)=2x^3+17x-8\)
Những câu hỏi liên quan
giai phuong trinh \(\sqrt{2x^2+11x+19}+\sqrt{2x^2+5x+7}=3\left(x+2\right)\)
gợi ý nhé
nhận thấy 2x2+11x+19=2x2+5x+7+6(x+2)
đặt ẩn phụ: căn(2x2+5x+7) = a và 3(x+2)=b
=) pt căn(a2+2b)+a=b (=) b(b-2a-2)=0 rồi giải từng trường hợp
giai phuong phuong trinh : \(\sqrt{2x^2+5x-3}-\sqrt{2x-1}=0\) 0
CM: \(|ab+cd|\subseteq\sqrt{\left(a^2+c^2\right)\left(b^2+d^2\right)}\)
\(\left(2x+3\right)\left(x+2\right)^2\left(2x+5\right)=3\)(Giai phuong trinh)
\(4\left(x^2+4x\right)^2+31\left(x^2+4x\right)+60=3\)
\(t=x^2+4x\)
\(4t^2+31t+57=0\)
\(\orbr{\begin{cases}t=\frac{-31-7}{8}=\frac{-19}{4}\\t=\frac{-31+7}{8}=-3\end{cases}}\)
\(x^2+4x+\frac{19}{4}=0\Rightarrow vn\)
\(x^2+4x+3=0\Rightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}\)
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Xem thêm câu trả lời
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)\(\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-1\)
1) giai phuong trinh:
a) \(x+\sqrt{2x+3}=2x\left(x-2\right)\)
Lời giải:
ĐK: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow \sqrt{2x+3}=2x^2-5x$
$\Leftrightarrow \sqrt{2x+3}-3=2x^2-5x-3$
$\Leftrightarrow \frac{2(x-3)}{\sqrt{2x+3}+3}=(2x+1)(x-3)$
$\Leftrightarrow (x-3)\left[\frac{2}{\sqrt{2x+3}+3}-(2x+1)\right]=0$
Xảy ra 2 TH:
TH1: $x-3=0\Rightarrow x=3$ (thỏa mãn)
TH2: $\frac{2}{\sqrt{2x+3}+3}=2x+1$
Đặt $\sqrt{2x+3}=t(t\geq 0)$ thì pt trở thành: \frac{2}{t+3}=t^2-2$
$\Leftrightarrow 2=(t^2-2)(t+3)\Leftrightarrow t^3+3t^2-2t-8=0$
$\Leftrightarrow (t+2)(t^2+t-4)=0$
Do $t\geq 0$ nên $t=\frac{-1+\sqrt{17}}{2}$
$\Leftrightarrow \sqrt{2x+3}=\frac{-1+\sqrt{17}}{2}\Leftrightarrow x=\frac{3-\sqrt{17}}{4}$ (thỏa mãn)
Vậy........
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Giai phuong trinh
1/ sqrt{x^2+4x+5}+sqrt{x^2-6x+13}3
2/ sqrt{3x^2-18x+28}+sqrt{4x^2-24x+45}6x-x^2-5
3/ sqrt{2x^2-4x+27}+sqrt{3x^2-6x+12}4x^2+8x+4
4/ sqrt{x^2+x+7}+sqrt{x^2+x+2}sqrt{3x^2+3x+19}
5/ left(x+2right)left(x+3right)-sqrt{x^2+5x+1}9
6/ left(x+4right)left(x+1right)-3sqrt{x^2+5x+2}6
7/ sqrt{2x^2+3x+5}+sqrt{2x^2-3x+5}3sqrt{x}
Đọc tiếp
Giai phuong trinh
1/ \(\sqrt{x^2+4x+5}+\sqrt{x^2-6x+13}=3\)
2/ \(\sqrt{3x^2-18x+28}+\sqrt{4x^2-24x+45}=6x-x^2-5\)
3/ \(\sqrt{2x^2-4x+27}+\sqrt{3x^2-6x+12}=4x^2+8x+4\)
4/ \(\sqrt{x^2+x+7}+\sqrt{x^2+x+2}=\sqrt{3x^2+3x+19}\)
5/ \(\left(x+2\right)\left(x+3\right)-\sqrt{x^2+5x+1}=9\)
6/ \(\left(x+4\right)\left(x+1\right)-3\sqrt{x^2+5x+2}=6\)
7/ \(\sqrt{2x^2+3x+5}+\sqrt{2x^2-3x+5}=3\sqrt{x}\)
Em xin phép làm bài EZ nhất :)
4,ĐK :\(\forall x\in R\)
Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))
\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)
\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)
\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy ....
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Giai phuong trinh giup minh 3 cau nay voi
a,\(3x\left(2-\sqrt{4}\right)=3\left(\sqrt{4}x+1\right)\)
b,\(\left(5-x\right).\left(\sqrt{3}+x\right)-5=0.\)
c,\(\left(x^2-2x\right)+\left(-4+8x\right)=0.\)
giai phuong trinh
x*(2x+3)2 -4x2+9=0
\(x-\sqrt{x-3}-5=0\)
\(x^3-4x^2-3x+6=0\)
\(3x^3+4x^2-5x-6=0\)
\(\sqrt{x^2+4x+8}+\sqrt{x^2+4x+4}=\sqrt{2\cdot\left(x^2+4x+6\right)}\)
Tưởng bn lớp 5 ạ?? Sao lại đăng câu hỏi lp 9 ạ??:)
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Ta có : x(2x + 3)2 - 4x2 + 9 = 0
<=> x(2x + 3)2 - (4x2 - 9) = 0
<=> x(2x + 3)2 - (2x - 3)(2x + 3) = 0
<=> (2x + 3)[x(2x + 3) - 2x + 3] = 0
<=> (2x + 3)(2x2 + 3x - 2x + 3) = 0
<=> (2x + 3)(2x2 + x + 3) = 0
<=> 2x + 3 = 0 (vì 2x2 + x + 3 > 0 với mọi x)
<=> 2x = -3
<=> x = -3/2
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Giai phuong trinh:
a)\(\frac{4+9x}{9x^21}=\frac{3}{3x+1}-\frac{2}{1-3x}\)
b)\(\frac{2x-3}{x+1}+\frac{x^2-5x+10}{\left(x+1\right)\left(x-3\right)}=\frac{3x-5}{x-3}\)
c)\(\frac{x\left(x+4\right)}{2x-3}=\frac{x^2+4}{2x-3}+1-\frac{2}{3-2x}\)
d)\(\frac{1}{x+2}+\frac{x}{x-3}=1-\frac{5x}{\left(x+2\right)\left(3-x\right)}-\frac{1}{x+2}\)
