giai phuong trinh \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}-\sqrt[3]{4x-3}=0\)
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giai phuong trinh: \(\sqrt[3]{x^2+4x+3}+\sqrt[3]{4x^2-9x-3}=\sqrt[3]{3x^2-2x+2}+\sqrt[3]{2x^2-3x-2}\)
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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minh lop 5 dang chi minh muon nick cua minh
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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
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: \(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2+2x+3}+\sqrt{x^2-x-1}\)
Giai phuong trinh
a/ \(\sqrt{4x^2+4x+1}\) - \(\sqrt{25x^2+10x+1}\) = 0
b/ \(\sqrt{x^4-16x^2+64}=\sqrt{25x^2+10x+1}\)
c/ \(\sqrt{x^2-25}-\sqrt{x-5}=0\)
d/ \(\sqrt{4x^2-9}-2\sqrt{2x+3}=0\)
e/ \(\sqrt{x-2}-3\sqrt{x^2-4}=0\)
a.
\(\sqrt{4x^2+4x+1}-\sqrt{25x^2+10x+1}=0\)
\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}-\sqrt{\left(5x+1\right)^2}=0\)
\(\Leftrightarrow2x+1-\left(5x+1\right)=0\)
\(\Leftrightarrow-3x=0\Leftrightarrow x=0\)
b.
\(\sqrt{x^4-16x^2+64}=\sqrt{25x^2+10x+1}\)
\(\Leftrightarrow\sqrt{\left(x^2-8\right)^2}=\sqrt{\left(5x+1\right)^2}\)
\(\Leftrightarrow x^2-8=5x+1\)
\(\Leftrightarrow x^2-5x+\dfrac{25}{4}=\dfrac{61}{4}\)
\(\Leftrightarrow\left(x-\dfrac{5}{2}\right)^2=\dfrac{61}{4}\)
............................
tương tự ..
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c: \(\Leftrightarrow\sqrt{x-5}\left(\sqrt{x+5}-1\right)=0\)
=>x-5=0 hoặc x+5=1
=>x=-4 hoặc x=5
d: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)
=>2x+3=0 hoặc 2x-3=4
=>x=7/2 hoặc x=-3/2
e: \(\Leftrightarrow\sqrt{x-2}\left(1-3\sqrt{x+2}\right)=0\)
=>x-2=0 hoặc 3 căn x+2=1
=>x=2 hoặc x+2=1/9
=>x=-17/9 hoặc x=2
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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
\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-2\sqrt{2x-5}}=2\sqrt{2}\)
GIAI PHUONG TRINH:
\(2\left(\sqrt{\frac{x-1}{4}-3}\right)=2\sqrt{\frac{4x-4}{9}}-\frac{1}{3}\)
Giai phương trình: \(\sqrt[3]{3x+1}+\sqrt[3]{5-x}+\sqrt[3]{2x-9}=\sqrt[3]{4x-3}\)
\(\Leftrightarrow\sqrt[3]{3x+1}+\sqrt[3]{2x-9}=\sqrt[3]{4x-3}+\sqrt[3]{x-5}=k\)
\(\Leftrightarrow5x-8+3\sqrt[3]{\left(3x+1\right)\left(2x-9\right)}.k=5x-8+3\sqrt[3]{\left(4x-3\right)\left(x-5\right)}.k\)
\(\Leftrightarrow\left[{}\begin{matrix}k=0\\\left(3x+1\right)\left(2x-9\right)=\left(4x-3\right)\left(x-5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=9-2x\\6x^2-25x-9=4x^2-23x+15\end{matrix}\right.\)