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Mk xin lỗi nha, câu c sai đề
c) (x+6)4 + (x+8)4 = 272
a) ĐKXĐ: x khác 0
\(x+\dfrac{5}{x}>0\)
\(\Leftrightarrow x^2+5>0\) ( luôn đúng)
Vậy bất pt vô số nghiệm ( loại x = 0)
d)
\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2-x-3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{-5}{8}\)
\(\Leftrightarrow2x+2-4x+4>-15\)
\(\Leftrightarrow-2x>-21\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
Vậy....................
a)\(x+\dfrac{5}{x}>0\left(ĐKXĐ:x\ne0\right)\)
\(\Leftrightarrow\dfrac{x^2+5}{x}>0\)
Mà \(x^2+5>0\)
\(\Rightarrow x>0\)
d)\(\dfrac{x+1}{12}-\dfrac{x-1}{6}>\dfrac{x-2}{8}-\dfrac{x+3}{8}\)
\(\Leftrightarrow\dfrac{x+1}{12}-\dfrac{2x-2}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow\dfrac{-x+3}{12}>\dfrac{-5}{8}\)
\(\Leftrightarrow-x+3>-\dfrac{15}{2}\)
\(\Leftrightarrow-x>-\dfrac{21}{2}\)
\(\Leftrightarrow x< \dfrac{21}{2}\)
a) \(4.\left(x-1\right)^2-9=0\)
\(\Rightarrow4.\left(x-1\right)^2=9\)
\(\Rightarrow\left(x-1\right)^2=9:4=\dfrac{9}{4}=\left(\pm\dfrac{3}{2}\right)^2\)
\(\Rightarrow x-1=\pm\dfrac{3}{2}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{3}{2}\\x-1=\dfrac{-3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
vậy\(\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
b) \(\dfrac{1}{4}-9.\left(x-1\right)^2=0\)
\(\Rightarrow9.\left(x-1\right)^2=\dfrac{1}{4}\)
\(\Rightarrow\left(x-1^2\right)=\dfrac{1}{36}=(\pm\dfrac{1}{6})^2\)
\(\Rightarrow x-1=\pm\dfrac{1}{6}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=\dfrac{1}{6}\\x-1=\dfrac{-1}{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=\dfrac{5}{6}\end{matrix}\right.\)
e) \(\dfrac{1}{16}-\left(2x+\dfrac{3}{4}\right)^2=0\)
\(\Rightarrow\left(2x+\dfrac{3}{4}\right)^2=\dfrac{1}{16}=\left(\pm\dfrac{1}{4}\right)^2\)
\(\Rightarrow2x+\dfrac{3}{4}=\pm\dfrac{1}{4}\)
\(\Rightarrow\)\(\left[{}\begin{matrix}2x+\dfrac{3}{4}=\dfrac{1}{4}\\2x+\dfrac{3}{4}=\dfrac{-1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
vậy \(\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-1}{2}\end{matrix}\right.\)
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
a) (x + 5)2 - (x - 3)2 = 2x - 7
(x + 5 - x + 3)(x + 5 + x - 3) = 2x - 7
8(2x + 2)= 2x - 7
16x + 16 = 2x - 7
16x - 2x = - 7 - 16
14x = - 23
x = - 23/14
b) (2x - 3)(4x2 + 6x + 9) = 98
(2x)3 - 33 = 98
8x3 - 27 = 98
8x3 = 125
x3 = 125/8
x3 = (5/2)3
x = 5/2
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
Vì dài quá nên mình chỉ có thể trả lời được mấy câu thôi
Bài 1:
27x3 - 8 : (6x + 9x2 +4)
= (3x - 2) (9x2 + 6x + 4) : (9x2 + 6x + 4)
= 3x - 2
Bài 3:
a, 81x4 + 4 = (9x2)2 + 36x2 + 4 - 36x2
= (9x2 + 2)2 - (6x)2
= (9x2 + 6x + 2)(9x2 - 6x + 2)
b, x2 + 8x + 15 = x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c, x2 - x - 12 = x2 + 3x - 4x - 12
= x(x + 3) - 4(x + 3)
= (x + 3) (x - 4)
Câu 1:
(27x3 - 8) : (6x + 9x2 + 4)
= (3x - 2)(9x2 + 6x + 4) : (6x + 9x2 + 4)
= 3x - 2
Câu 2:
a) (3x - 5)(2x+ 11) - (2x + 3)(3x + 7)
= 6x2 + 33x - 10x - 55 - 6x2 - 14x - 9x - 21
= -76
⇒ đccm
b) (2x + 3)(4x2 - 6x + 9) - 2(4x3 - 1)
= 8x3 + 27 - 8x3 + 2
= 29
⇒ đccm
Câu 3:
a) 81x4 + 4
= (9x2)2 + 22
= (9x2 + 2)2 - (6x)2
= (9x2 - 6x + 2)(9x2 + 6x + 2)
b) x2 + 8x + 15
= x2 + 3x + 5x + 15
= x(x + 3) + 5(x + 3)
= (x + 3)(x + 5)
c) x2 - x - 12
= x2 - 4x + 3x - 12
= x(x - 4) + 3(x - 4)
= (x - 4)(x + 3)
Bài 1:
a) \(\left(x+2\right)^2-x^2+4=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+4\right)=0\)
\(\Leftrightarrow x+2=0\) hoặc \(x+4=0\)
\(\Leftrightarrow x=-2\) hoặc \(x=-4\)
b) \(2x^3+\dfrac{3}{2}x^2=0\)
\(\Leftrightarrow x^2\left(2x+\dfrac{3}{2}\right)=0\)
\(\Leftrightarrow x^2=0\) hoặc \(2x+\dfrac{3}{2}=0\)
\(\Leftrightarrow x=0\) hoặc \(x=-\dfrac{3}{4}\)
bài 1
a) (x+2)2-x2+4=0
\(\Leftrightarrow\)x2+4x+4-x2+4=0
\(\Leftrightarrow\)4x+8=0
\(\Leftrightarrow\) 4(x+2)=0
=>x+2=0
\(\Leftrightarrow\)x=-2
vậy x=-2
b) \(2x^3+\dfrac{3}{2}x^2=0\)
\(\Leftrightarrow x^2\left(2x+\dfrac{3}{2}\right)=0\)
=>\(\left[{}\begin{matrix}x^2=0\\2x+\dfrac{3}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\-\dfrac{3}{4}\end{matrix}\right.\)
vậy x=0 hoặc x=-\(\dfrac{3}{4}\)
Bài 4 : Tìm x biết:
a, 4x2 - 49 = 0
\(\Leftrightarrow\) (2x)2 - 72 = 0
\(\Leftrightarrow\) (2x - 7)(2x + 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}2x-7=0\\2x+7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
b, x2 + 36 = 12x
\(\Leftrightarrow\) x2 + 36 - 12x = 0
\(\Leftrightarrow\) x2 - 2.x.6 + 62 = 0
\(\Leftrightarrow\) (x - 6)2 = 0
\(\Leftrightarrow\) x = 6
e, (x - 2)2 - 16 = 0
\(\Leftrightarrow\) (x - 2)2 - 42 = 0
\(\Leftrightarrow\) (x - 2 - 4)(x - 2 + 4) = 0
\(\Leftrightarrow\) (x - 6)(x + 2) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=-2\end{matrix}\right.\)
f, x2 - 5x -14 = 0
\(\Leftrightarrow\) x2 + 2x - 7x -14 = 0
\(\Leftrightarrow\) x(x + 2) - 7(x + 2) = 0
\(\Leftrightarrow\) (x + 2)(x - 7) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)
a)(x+1)(x2+2x)=(x+1)x(x+2)=0
\(=>\left\{{}\begin{matrix}x+1=0=>x=-1\\x=0\\x+2=0=>x=-2\end{matrix}\right.\)
b)x(3x-2)-5(2-3x)=x(3x-2)+5(3x-2)=(3x-2)(x+5)=0
\(=>\left\{{}\begin{matrix}3x-2=0=>x=\dfrac{2}{3}\\x+5=0=>x=-5\end{matrix}\right.\)
c)\(\dfrac{4}{9}-25x^2=\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)\)
=0
\(=>\left\{{}\begin{matrix}\dfrac{2}{3}-5x=0=>x=\dfrac{2}{15}\\\dfrac{2}{3}+5x=0=>x=\dfrac{-2}{15}\end{matrix}\right.\)
d)\(x^2-x+\dfrac{1}{4}=x^2-2.\dfrac{1}{2}.x+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2=0\)
\(=>x-\dfrac{1}{2}=0=>x=\dfrac{1}{2}\)
(a) \(9x^2+12x+4=0\)
\(\Leftrightarrow\left(3x+2\right)^2=0\Leftrightarrow3x+2=0\Leftrightarrow x=-\dfrac{3}{2}\)
(b) \(x^2+\dfrac{1}{4}=x\)
\(\Leftrightarrow x^2-x+\dfrac{1}{4}=0\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\Leftrightarrow x-\dfrac{1}{2}=0\Leftrightarrow x=\dfrac{1}{2}\)
(c) \(4-\dfrac{12}{x}+\dfrac{9}{x^2}=0\left(x\ne0\right)\)
\(\Leftrightarrow\left(2-\dfrac{3}{x}\right)^2=0\Leftrightarrow2-\dfrac{3}{x}=0\Leftrightarrow x=\dfrac{3}{2}\)