Tìm x, biết :
a) \(\dfrac{2}{3}x\left(x^2-4\right)=0\)
b) \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
c) \(\left(x^2-y^2+6x+9\right):\left(x+y+3\right)\)
Những câu hỏi liên quan
Tìm x, biết : a/ dfrac{1}{3}xleft(x^2-4right)0b/ xleft(x+5right)x+5c/ x^3-dfrac{1}{9}x03)^2-left(x+5right)^20e/ left(x+2right)^2-left(x-2right)left(x+2right)0f/ xleft(2x-3right)-6+4x0g/ 2left(3x-2right)^2-9x^2+40h/ x^2left(x+1right)+2xleft(x+1right)0i/ 4x^2+9x+50
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Tìm x, biết :
a/ \(\dfrac{1}{3}x\left(x^2-4\right)=0\)
b/ \(x\left(x+5\right)=x+5\)
c/ \(x^3-\dfrac{1}{9}x=0\)
3)\(^2-\left(x+5\right)^2=0\)
e/ \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
f/ \(x\left(2x-3\right)-6+4x=0\)
g/ \(2\left(3x-2\right)^2-9x^2+4=0\)
h/ \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
i/ \(4x^2+9x+5=0\)
a) \(\Rightarrow\dfrac{1}{3}x\left(x-2\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x+5\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
c) \(\Rightarrow x\left(x^2-\dfrac{1}{9}\right)=0\Rightarrow x\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{1}{3}\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
e) \(\Rightarrow\left(x+2\right)\left(x+2-x+2\right)=0\Rightarrow\left(x+2\right).4=0\Rightarrow x=-2\)
f) \(\Rightarrow x\left(2x-3\right)+2\left(2x-3\right)=0\Rightarrow\left(2x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
g) \(\Rightarrow2\left(3x-2\right)^2-\left(3x-2\right)\left(3x+2\right)=0\Rightarrow\left(3x-2\right)\left(3x-6\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=2\end{matrix}\right.\)
h) \(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
i) \(\Rightarrow4x\left(x+1\right)+5\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(4x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{4}\end{matrix}\right.\)
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a:dfrac{b}{left(a-4right)^2}.sqrt{dfrac{left(a-4right)^4}{b^2}}left(b0;ane4right)
b:dfrac{xsqrt{x}-ysqrt{y}}{sqrt{x}-sqrt{y}}left(xge0;yge0;xne0right)
c:dfrac{a}{left(b-2right)^2}.sqrt{dfrac{left(b-2right)^4}{a^2}left(a0;bne2right)}
d:dfrac{x}{left(y-3right)^2}.sqrt{dfrac{left(y-3right)^2}{x^2}left(x0;yne3right)}
e:2x +dfrac{sqrt{1-6x+9x^2}}{3x-1}
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a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
a, \(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}=\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}=1\)
b, Đặt \(B=\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\sqrt{x}=a,\sqrt{y}=b\)
Ta có: \(B=\dfrac{a^3-b^3}{a-b}=\dfrac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a-b}=a^2+ab+b^2\)
\(\Rightarrow B=x+\sqrt{xy}+y\)
Vậy...
c, \(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}}=\dfrac{a}{\left(b-2\right)^2}.\dfrac{\left(b-2\right)^2}{a}=1\)
d, \(2x+\dfrac{\sqrt{1-6x+9x^2}}{3x-1}=2x+\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}=2x+1\)
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a:b(a−4)2.√(a−4)4b2(b>0;a≠4)b(a−4)2.(a−4)4b2(b>0;a≠4)
= \(\dfrac{b}{\left(a-4\right)}.\dfrac{\sqrt{\left[\left(a-4\right)^2\right]^2}}{\sqrt{b^2}}\)
=\(\dfrac{b}{\left(a-4\right)^2}.\dfrac{\left(a-4\right)^2}{b}\)
= 1 ( nhân tử với tử mẫu với mẫu rồi rút gọn)
b:x√x−y√y√x−√y(x≥0;y≥0;x≠0)xx−yyx−y(x≥0;y≥0;x≠0)
=\(\dfrac{\sqrt{x^3}-\sqrt{y^3}}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\sqrt{x}-\sqrt{y}}\)
=\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right).\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}\)(áp dụng hằng đẳng thức )
= (x+\(\sqrt{xy}\)+y)
c:a(b−2)2.√(b−2)4a2(a>0;b≠2)a(b−2)2.(b−2)4a2(a>0;b≠2)
Tương tự câu a
d:x(y−3)2.√(y−3)2x2(x>0;y≠3)x(y−3)2.(y−3)2x2(x>0;y≠3)
tương tự câu a
e:2x +√1−6x+9x23x−1
= \(2x+\dfrac{\sqrt{\left(3x\right)^2-6x+1}}{3x-1}\)
= 2x+\(\dfrac{\sqrt{\left(3x-1\right)^2}}{3x-1}\)(hằng đẳng thức)
=2x+\(\dfrac{3x-1}{3x-1}\)
=2x+1
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5 tháng 10 2021 lúc 20:58
a) làm tính chia
\(\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]:\left(y-x\right)^2\)
b) tìm \(x\)
\(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)
ghi chú: đừng làm tắt được ko ạ?
b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)
\(\Leftrightarrow-4x+3+5x+2=0\)
\(\Leftrightarrow x=-5\)
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Tìm x,biết
a)\(\left(x-2^2\right)-1=0\)
b)\(4-\left(x-2\right)^2=0\)
c)\(x^2-9-\dfrac{8}{9}x^2=0\)
d)\(\left(3x-2\right)^2-\left(2x+3\right)^2=5\left(x+4\right)\left(x-4\right)\)
a. (x - 22) - 1 = 0
<=> x - 4 - 1 = 0
<=> x = 5
b. 4 - (x - 2)2 = 0
<=> 22 - (x - 2)2 = 0
<=> (2 - x + 2)(2 + x - 2) = 0
<=> x(4 - x) = 0
<=> \(\left[{}\begin{matrix}x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
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d. (3x - 2)2 - (2x + 3)2 = 5(x + 4)(x - 4)
<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x2 - 16)
<=> (x - 5)(5x + 1) = 5x2 - 80
<=> 5x2 + x - 25x - 5 = 5x2 - 80
<=> 5x2 - 5x2 + x - 25x = -80 + 5
<=> -24x = -75
<=> x = \(\dfrac{25}{8}\)
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a)\(\left(x-2^2\right)-1=0\Rightarrow x-4-1=0\Rightarrow x=5\)
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Xem thêm câu trả lời
Rút gọn:a) dfrac{3left(x-yright)left(x-zright)^2}{6left(x-yright)left(x-zright)}b) dfrac{6x^2y^2}{8xy^5}c) dfrac{3xleft(1-xright)}{2left(x-1right)}d) dfrac{9-left(x+5right)^2}{x^2+4x+4}e) dfrac{x^2-2x+1}{x^2-1}f) dfrac{8x-4}{8x^3-1}g) dfrac{x^2+5x+6}{x^2+4x+4}k) dfrac{20x^2-45}{left(2x+3right)^2}
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Rút gọn:
a) \(\dfrac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}\)
b) \(\dfrac{6x^2y^2}{8xy^5}\)
c) \(\dfrac{3x\left(1-x\right)}{2\left(x-1\right)}\)
d) \(\dfrac{9-\left(x+5\right)^2}{x^2+4x+4}\)
e) \(\dfrac{x^2-2x+1}{x^2-1}\)
f) \(\dfrac{8x-4}{8x^3-1}\)
g) \(\dfrac{x^2+5x+6}{x^2+4x+4}\)
k) \(\dfrac{20x^2-45}{\left(2x+3\right)^2}\)
a: \(=\dfrac{x-z}{2}\)
b: \(=\dfrac{3x}{4y^3}\)
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Bài 1: Tính: a)dfrac{x+y}{2left(x-yright)}-dfrac{x-y}{2left(x+yright)}-dfrac{2y^2}{y^2-x^2} b)left(dfrac{9}{x^3-9x}+dfrac{1}{x+3}right):left(dfrac{x-3}{x^2+3}-dfrac{x}{3x+9}right)Bài 2: Tìm x: a)2x^3-50x0 b)x^3+x^2+x+a chia hết cho x+1Bài 3: Cho △MNP vuông tại N, biết MN 6cm, NP 8cm. đường cao NH, qua H kẻ HC⊥MN, HD⊥NP a) Chứng minh HDNC là hình chữ nhật. b) Tính CD c) Tính diện tích △NMH
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Bài 1: Tính:
a)\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}-\dfrac{2y^2}{y^2-x^2}\)
b)\(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3}-\dfrac{x}{3x+9}\right)\)
Bài 2: Tìm x:
a)2x\(^3\)-50x=0 b)\(x^3+x^2+x+a\) chia hết cho x+1
Bài 3: Cho △MNP vuông tại N, biết MN = 6cm, NP = 8cm. đường cao NH, qua H kẻ HC⊥MN, HD⊥NP
a) Chứng minh HDNC là hình chữ nhật.
b) Tính CD
c) Tính diện tích △NMH
Bài 1:
\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)
Bài 2:
\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)
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Bài 1: Cm giá trị của biểu thức sau không phụ thuộc vào biến x:
a) left(2x+3right).left(4x^2-6x+9right)-2left(4x^3-1right)
b)left(x+3right)^3+left(x+9right).left(x^2+27right)
c)left(x+yright).left(x^2-xy+y^2right)+left(x-yright).left(x^2+xy+y^2right)-2x^3
Bài 2: Tìm x biết:
a) left(x+2right)^2-90
b) left(x+2right)^2-x^2+40
c) left(x-3right)^2-40
d) x^2-2x24
e) left(2x-1right)^2+left(x+3right)^2-5left(x-7right)left(x+7right)0
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Bài 1: Cm giá trị của biểu thức sau không phụ thuộc vào biến x:
a) \(\left(2x+3\right).\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
b)\(\left(x+3\right)^3+\left(x+9\right).\left(x^2+27\right)\)
c)\(\left(x+y\right).\left(x^2-xy+y^2\right)+\left(x-y\right).\left(x^2+xy+y^2\right)-2x^3\)
Bài 2: Tìm x biết:
a) \(\left(x+2\right)^2-9=0\)
b) \(\left(x+2\right)^2-x^2+4=0\)
c) \(\left(x-3\right)^2-4=0\)
d) \(x^2-2x=24\)
e) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
Bài 1:
a) \(\left(2x+3\right)\cdot\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3-12x^2+18x+12x^2-18x+27-8x^3-3=27-3=24\)
--> đpcm
b) Sửa đề: \(\left(x+3\right)^3-\left(x+9\right)\left(x^2+27\right)\)
\(=x^3+9x^2+27x+27-\left(x^3+27x+9x^2+243\right)\)
\(=x^3+9x^2+27x+27-x^3-27x-9x^2-243=27-243=-216\)
--> đpcm
c) \(\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x-y\right)\left(x^2+xy+y^2\right)-2x^3\)
\(=x^3+y^3+x^3-y^3-2x^3=2x^3-2x^3=0\)
--> đpcm
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B1: a) \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3-27-8x^3+2\)
\(=-25\)
b) c) Làm theo câu a áp dụng HĐT.
B2:
a) \(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2+3\right)\left(x+2-3\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-5\\x=1\end{matrix}\right..\)
Mấy câu b,c,d bn chịu khó tạo HĐT nhé.
e) \(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=-\dfrac{255}{2}\)
Vậy \(x=-\dfrac{255}{2}\)
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bài 2
a) ( x+2)2-9=0
=>(x+2)2- 32 =0
=> (x+2-3)(x+2+3)=0
=>(x-1)(x+5)=0
=>\(\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
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Xem thêm câu trả lời
tính đạo hàm
a) \(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}\)
b) \(y=x+3+\dfrac{4}{x+3}\) giải phương trình y'=0
c) \(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\) tính y'(-1)
d) \(y=x-2+\dfrac{9}{x-2}\) giải phương trình y'=0
a:
ĐKXĐ: \(x\notin\left\{\dfrac{3}{2};1\right\}\)
\(y=\dfrac{\left(x-2\right)^2}{\left(2x-3\right)\left(x-1\right)}=\dfrac{x^2-4x+4}{2x^2-2x-3x+3}\)
=>\(y=\dfrac{x^2-4x+4}{2x^2-5x+3}\)
=>\(y'=\dfrac{\left(x^2-4x+4\right)'\left(2x^2-5x+3\right)-\left(x^2-4x+4\right)\left(2x^2-5x+3\right)'}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{\left(2x-4\right)\left(2x^2-5x+3\right)-\left(2x-5\right)\left(x^2-4x+4\right)}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{4x^3-10x^2+6x-8x^2+20x-12-2x^3+8x^2-8x+5x^2-20x+20}{\left(2x^2-5x+3\right)^2}\)
=>\(y'=\dfrac{2x^3-5x^2-2x+8}{\left(2x^2-5x+3\right)^2}\)
b:
ĐKXĐ: x<>-3
\(y=\left(x+3\right)+\dfrac{4}{x+3}\)
=>\(y'=\left(x+3+\dfrac{4}{x+3}\right)'=1+\left(\dfrac{4}{x+3}\right)'\)
\(=1+\dfrac{4'\left(x+3\right)-4\left(x+3\right)'}{\left(x+3\right)^2}\)
=>\(y'=1+\dfrac{-4}{\left(x+3\right)^2}=\dfrac{\left(x+3\right)^2-4}{\left(x+3\right)^2}\)
y'=0
=>\(\left(x+3\right)^2-4=0\)
=>\(\left(x+3+2\right)\left(x+3-2\right)=0\)
=>(x+5)(x+1)=0
=>x=-5 hoặc x=-1
c:
ĐKXĐ: x<>-2
\(y=\dfrac{\left(5x-1\right)\left(x+1\right)}{x+2}\)
=>\(y=\dfrac{5x^2+5x-x-1}{x+2}=\dfrac{5x^2+4x-1}{x+2}\)
=>\(y'=\dfrac{\left(5x^2+4x-1\right)'\left(x+2\right)-\left(5x^2+4x-1\right)\left(x+2\right)'}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{\left(5x+4\right)\left(x+2\right)-\left(5x^2+4x-1\right)}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{5x^2+10x+4x+8-5x^2-4x+1}{\left(x+2\right)^2}\)
=>\(y'=\dfrac{10x+9}{\left(x+2\right)^2}\)
\(y'\left(-1\right)=\dfrac{10\cdot\left(-1\right)+9}{\left(-1+2\right)^2}=\dfrac{-1}{1}=-1\)
d:
ĐKXĐ: x<>2
\(y=x-2+\dfrac{9}{x-2}\)
=>\(y'=\left(x-2+\dfrac{9}{x-2}\right)'=1+\left(\dfrac{9}{x-2}\right)'\)
\(=1+\dfrac{9'\left(x-2\right)-9\left(x-2\right)'}{\left(x-2\right)^2}\)
=>\(y'=1+\dfrac{-9}{\left(x-2\right)^2}=\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}\)
y'=0
=>\(\dfrac{\left(x-2\right)^2-9}{\left(x-2\right)^2}=0\)
=>\(\left(x-2\right)^2-9=0\)
=>(x-2-3)(x-2+3)=0
=>(x-5)(x+1)=0
=>x=5 hoặc x=-1
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Bài 2:a. 2x^2+2xy+y^2+96x-left|y+3right| Leftrightarrowleft|y+3right|6x-2x^2-2xy-y^2-9 Leftrightarrowleft|y+3right|-x^2-2xy-y^2-x^2+6x-9 Leftrightarrowleft|y+3right|-left(x+yright)^2-left(x-3right)^2 Leftrightarrowleft|y+3right|-left[left(x+yright)^2+left(x-3right)^2right] Có: left|y+3right|ge0 -left[left(x+yright)^2+left(x-3right)^2right]le0 Do đó: left|y+3right|-left[left(x+yright)^2+left(x-3right)^2right]0 Leftrightarrowhept{begin{cases}y+30x+y0x-30end{cases}}Leftrightarrowhept{begin{cases}x3...
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Bài 2:
a. \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)
\(\Leftrightarrow\left|y+3\right|=6x-2x^2-2xy-y^2-9\)
\(\Leftrightarrow\left|y+3\right|=-x^2-2xy-y^2-x^2+6x-9\)
\(\Leftrightarrow\left|y+3\right|=-\left(x+y\right)^2-\left(x-3\right)^2\)
\(\Leftrightarrow\left|y+3\right|=-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]\)
Có: \(\left|y+3\right|\ge0\)
\(-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]\le0\)
Do đó: \(\left|y+3\right|=-\left[\left(x+y\right)^2+\left(x-3\right)^2\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}y+3=0\\x+y=0\\x-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)
b. \(\left(2x^2+x-2013\right)^2+4\left(x^2-5x-2012\right)^2=4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)\)
\(\Leftrightarrow\left(2x^2+x-2013\right)^2-4\left(2x^2+x-2013\right)\left(x^2-5x-2012\right)+\left[2\left(x^2-5x-2012\right)\right]^2=0\)
\(\Leftrightarrow\left(2x^2+x-2013-2x^2+10x+4024\right)^2=0\)
\(\Leftrightarrow\left(11x+2011\right)^2=0\)
\(\Leftrightarrow11x+2011=0\)
\(\Leftrightarrow x=-\frac{2011}{11}\)