P=/3x-5/+/x-2/
Q=/x-3/-2./-5x/
Những câu hỏi liên quan
a)sqrt{1-x}left(x-3x^2right)x^3-3x^2+2x+6
b)x^2+x+12sqrt{x+1}36
c)3x-1+frac{x-1}{4x}sqrt{3x+1}
d)sqrt{x^2+12}-3xsqrt{x^2+5}-5
e)4x^2+12+sqrt{x-1}4left(xsqrt{5x-1}+sqrt{9-5x}right)
f)4x^3-25x^2+43x+xsqrt{3x-2}22+sqrt{3x-2}
g)2left(x+1right)sqrt{x}+sqrt{3left(2x^3+5x^2+4x+1right)}5x^3-3x^2+8
h)sqrt{x^2+12}-sqrt{x^2+5}3x-5
i)sqrt{1-3x}-sqrt[3]{3x-1}left|6x-2right|
k)sqrt{2x^3+3x^2-1}2x^2+2x-x^3-1
l)sqrt{x^2+x-2}+x^2sqrt{2left(x-1right)}+1
Đọc tiếp
a)\(\sqrt{1-x}\left(x-3x^2\right)=x^3-3x^2+2x+6\)
b)\(x^2+x+12\sqrt{x+1}=36\)
c)\(3x-1+\frac{x-1}{4x}=\sqrt{3x+1}\)
d)\(\sqrt{x^2+12}-3x=\sqrt{x^2+5}-5\)
e)\(4x^2+12+\sqrt{x-1}=4\left(x\sqrt{5x-1}+\sqrt{9-5x}\right)\)
f)\(4x^3-25x^2+43x+x\sqrt{3x-2}=22+\sqrt{3x-2}\)
g)\(2\left(x+1\right)\sqrt{x}+\sqrt{3\left(2x^3+5x^2+4x+1\right)}=5x^3-3x^2+8\)
h)\(\sqrt{x^2+12}-\sqrt{x^2+5}=3x-5\)
i)\(\sqrt{1-3x}-\sqrt[3]{3x-1}=\left|6x-2\right|\)
k)\(\sqrt{2x^3+3x^2-1}=2x^2+2x-x^3-1\)
l)\(\sqrt{x^2+x-2}+x^2=\sqrt{2\left(x-1\right)}+1\)
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
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c,\(pt\Leftrightarrow3\left(x-1\right)+\frac{x-1}{4x}+\left(2-\sqrt{3x+1}\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}\right)=0\)
\(\Rightarrow x=1\)
\(3+\frac{1}{4x}+\frac{1}{2+\sqrt{3x+1}}=0\)
bạn làm nốt pần này nhá
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a) (2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)
b)(8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)
c)(\(^{x^2}\)-5)(x+2)+5x=2\(^{x^2}\)+17
d)(\(^{x^2}\)-x+1)(x+1)-\(x^3\)+3x=15
a: \(\Leftrightarrow2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
=>5x=22
hay x=22/5
b: \(\Leftrightarrow24x^2+16x-9x-6-4x^2-16x-7x-28=10x^2-2x+5x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
hay \(x\in\left\{3;-\dfrac{11}{10}\right\}\)
c: \(\Leftrightarrow x^3+2x^2-5x-10+5x=2x^2+17\)
\(\Leftrightarrow x^3+2x^2-10-2x^2-17=0\)
=>x3=27
=>x=3
d: \(\Leftrightarrow x^3+1-x^3+3x=15\)
=>3x=14
hay x=14/3
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cho các đa thức sau
f(x)=\(3x^4+3x^3_{_{_{_{ }}}}-5x^2+x-5\)
g(x)=\(x^4+3x^3-3x^2+5x-7\)
h(x)=\(5x^4+2x^3+x^2-5\)
tìm các đa thức P(x) vafQ(x) biết P(x)+Q(x)=f(x)-g(x) và P(x)-Q(x)=g(x)+h(x)
\(P\left(x\right)+Q\left(x\right)=f\left(x\right)-g\left(x\right)\)
\(f\left(x\right)-g\left(x\right)=3x^4+3x^3-5x^2+x-5-x^4-3x^3+3x^2-5x+7\)
\(=2x^4-2x^2-4x+2\)
\(\Rightarrow P\left(x\right)+Q\left(x\right)=2x^4-2x^2-4x+2\left(1\right)\)
\(P\left(x\right)-Q\left(x\right)=g\left(x\right)+h\left(x\right)\)
\(g\left(x\right)+h\left(x\right)=x^4+3x^3-3x^2+5x-7+5x^4+2x^3+x^2-5\)
\(=6x^4+5x^3-2x^2+5x-12\)
\(\Rightarrow P\left(x\right)-Q\left(x\right)=6x^4+5x^3-2x^2+5x-12\left(2\right)\)
Từ ( 1 );( 2 ) thì tìm dc P(x) và Q(x)
a,P(x)= -x(x+5)-(2x-3)+x^2(3x-2)
b,Q(x)=5x^2-2(x+1)+3x(x-2)+5
a: \(P=-x^2-5x-2x+3+3x^3-2x^2=3x^3-3x^2-7x+3\)
b: \(Q\left(x\right)=5x^2-2x-2+3x^2-6x+5=8x^2-8x+3\)
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\(x^2+5x+6=3x+3\cdot4+2x-9\)
\(2\sqrt{x}+8x+5=5x-4+3x+19\)
\(5\sqrt{x}+2x-8=5x+4-3x-19\)
\(2x^2+5z+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)
\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)
\(2\sqrt{3x}+11x-18=5x+2+6\cdot\sqrt{3x}+6x-21\)
1)2x(25x-4)-(5x-2)(5x+1)8 / 5)2left(x-2right)-3left(3x-1right)left(x-3right)
2)x(4x-3)-(2x-2)(2x-1)5 / 6)frac{2}{x+1}-frac{1}{x-2}frac{3x-11}{left(x+1right)left(x-2right)}
3)frac{5}{2x+3}+frac{3}{9-x^2}frac{8}{7left(x3right)} / 7)frac{5x-2}{6}+frac{3-4x}{2}2-frac{x+7}{3}
4)frac{2}{3left(x-2right)}+frac{5}{12-3x^2}frac{3}{4left(x+2right)} /...
Đọc tiếp
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
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1. (4x-10).(24+5x)=0
2 .(2x-5).(3x-2)=0
3. (2x-1).(3x+1)=0
4. x.(\(x\)2-1)=0
5.(5x+3).(\(x^2\)+4).(x-1)=0
6.(x-1).(x+2).(x+3)=0
7.(x-1).(x+5).(-3x+8)=0
a)
\(\left(4x-10\right)\cdot\left(24+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=-\frac{24}{5}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{5}{2};-\frac{24}{5}\right\}\)
b)
\(\left(2x-5\right)\left(3x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{5}{2};\frac{2}{3}\right\}\)
c)
\(\left(2x-1\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(S=\left\{\frac{1}{2};-\frac{1}{3}\right\}\)
d)
\(x\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(S=\left\{0;\frac{1}{2}\right\}\)
e) \(\left(5x+3\right)\left(x^2+4\right)\left(x-1\right)=0\)
Do \(x^2\ge0\) Nên \(x^2+4>0\)
\(\left(5x+3\right)\left(x^2+4\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{5}\\x=1\end{matrix}\right.\)
Vậy \(S=\left\{-\frac{3}{5};1\right\}\)
....... Còn lại cứ cho mỗi thừa số = 0 rồi tìm x như bình thường thôi bạn
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1. (4x - 10)(24 + 5x) = 0
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{-24}{5}\end{matrix}\right.\)
Vậy S = {\(\frac{5}{2}\); \(\frac{-24}{5}\)}
2. (2x - 5)(3x - 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)
Vậy S = {\(\frac{5}{2}\); \(\frac{2}{3}\)}
3. (2x - 1)(3x + 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{3}\end{matrix}\right.\)
Vậy S = {\(\frac{1}{2}\); \(\frac{-1}{3}\)}
4. x(x2 - 1) = 0
\(\Leftrightarrow\) x(x - 1)(x + 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy S = {0; 1; -1}
5. (5x + 3)(x2 + 4)(x - 1) = 0
VÌ x2 + 4 > 0 với mọi x nên
\(\Rightarrow\left[{}\begin{matrix}5x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{5}\\x=1\end{matrix}\right.\)
Vậy S = {\(\frac{-3}{5}\); 1}
6. (x - 1)(x + 2)(x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\x=-3\end{matrix}\right.\)
Vậy S = {1; -2; -3}
7. (x - 1)(x + 5)(-3x + 8) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+5=0\\-3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\\x=\frac{8}{3}\end{matrix}\right.\)
Vậy S = {1; -5; \(\frac{8}{3}\)}
Chúc bn học tốt!!
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Cho P(x) = x\(^2+5x^4-3x^3+x^2+4x^4+3x^3-x+5\)
Q(x)= x - 5x\(^3\)\(-x^2-x^4+4x^3-x^2+3x-1\)
a, Thu gọn, sắp xếp theo lũy thừa giảm của P(x)
b, Tính P(x) + Q(x); P(x) - Q(x)
a/ Thu gọn và sắp xếp:
\(P\left(x\right)=x^2+5x^4-3x^3+x^2+4x^4+3x^3-x+5=\left(5x^4+4x^4\right)+\left(3x^3-3x^3\right)+\left(x^2+x^2\right)-x+5=9x^4+2x^2-x+5\)
---
\(Q\left(x\right)=x-5x^3-x^2-x^4+4x^3-x^2+3x-1=-x^4+\left(4x^3-5x^3\right)+\left(-x^2-x^2\right)+\left(x+3x\right)-1=-x^4-x^3-2x^2+4x-1\)
b/ \(P\left(x\right)+Q\left(x\right)=9x^4+2x^2-x+5+\left(-x^4-x^3-2x^2+4x-1\right)=9x^4+2x^2-x+5-x^4-x^3-2x^2+4x-1=8x^4-x^3+3x+4\)
--
\(P\left(x\right)-Q\left(x\right)=9x^4+2x^2-x+5-\left(-x^4-x^3-2x^2+4x-1\right)=9x^4+2x^2-x+5+x^4+x^3+2x^2-4x+1=10x^4+x^3+4x^2-5x+6\)
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P(x) = x^2 + 5x^4 - 3x^3 + x^2 +4x^4 + 3x^3 - x + 5 cho đa thức q (x)= x - 5x^3 - x^2 - x^4 + 4x^3 - x^2 - 3x-1 thu gọn rồi sắp xếp các đa thức trên theo lũy thừa giảm dần của biến tính p (x )+ Q(x) và p( x )trừ Q(x)
Ta có: \(P\left(x\right)=x^2+5x^4-3x^3+x^2+4x^4+3x^3-x+5\)
\(=9x^4+2x^2-x+5\)
Ta có: \(Q\left(x\right)=x-5x^3-x^2-x^4+4x^3-x^2-3x-1\)
\(=-x^4-x^3-2x^2-2x-1\)
Ta có: P(x)+Q(x)
\(=9x^4+2x^2-x+5-x^4-x^3-2x^2-2x-1\)
\(=8x^4-x^3-3x+4\)
Ta có: P(x)-Q(x)
\(=9x^4+2x^2-x+5+x^4+x^3+2x^2+2x+1\)
\(=10x^4+x^3+4x^2+x+6\)
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Giải phương trình:
1) \(3x^2+4x-3=4x\sqrt{4x-3}\)
2) \(\sqrt{x^2+2x}+\sqrt{x+2}=\sqrt{x}+\sqrt{x^2+2x-2}\)
3) \(\sqrt{3x+8}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{5x-7}\)
4) \(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\)
1/ \(3x^2+4x-3=4x\sqrt{4x-3}\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{4x-3}+4x-3\right)-x^2=0\)
\(\Leftrightarrow\left(2x-\sqrt{4x-3}\right)^2-x^2=0\)
\(\Leftrightarrow\left(3x-\sqrt{4x-3}\right)\left(x-\sqrt{4x-3}\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x=\sqrt{4x-3}\\x=\sqrt{4x-3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}9x^2-4x+3=0\\x^2-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\x=3\end{matrix}\right.\)
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3.\(pt\Leftrightarrow\sqrt{3x+8}-\sqrt{3x+5}=\sqrt{5x-4}-\sqrt{5x-7}\)
\(\Leftrightarrow\frac{3x+8-5x+4}{\sqrt{3x+8}+\sqrt{5x+4}}-\frac{3x+5-5x+7}{\sqrt{3x+5}+\sqrt{5x+7}}=0\)
\(\Leftrightarrow\left(12-2x\right)\left(\frac{1}{\sqrt{3x+8}+\sqrt{5x+4}}+\frac{1}{\sqrt{3x+5}+\sqrt{5x+7}}\right)=0\)
\(\Rightarrow x=6\)
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