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làm nốt
d) (2x-1)(3x+2)(3-x)
=(6x2+x-2)(3-x)
=-6x3+17x2+5x-6
e) (x+3)(x2+3x-5)
=x3+6x2+4x-15
f) (xy-2)(x3-2x-6)
=x4y-2x3-2x2y-6xy+4x+12
g) (5x3-x2+2x-3)(4x2-x+2)
=20x5-9x4+19x3-16x2+7x-6
Bài 1:
a) (x-2)(x2+3x+4)
=x(5x+4)-2(5x+4)
= 5x2+4x-10x-8
=5x2-6x-8
a: (x-3)(x-2)<0
=>x-2>0 và x-3<0
=>2<x<3
b: \(\left(x+3\right)\left(x+4\right)\left(x^2+2\right)\ge0\)
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\ge0\)
=>x>=-3 hoặc x<=-4
c: \(\dfrac{x-1}{x-2}\ge0\)
nên \(\left[{}\begin{matrix}x-2>0\\x-1\le0\end{matrix}\right.\Leftrightarrow x\in(-\infty;1]\cup\left(2;+\infty\right)\)
d: \(\dfrac{x+3}{2-x}\ge0\)
\(\Leftrightarrow\dfrac{x+3}{x-2}\le0\)
hay \(x\in[-3;2)\)
a) x3 + 3x2 + 3x + 1 = 64
=> (x + 1)3 = 64
=> (x + 1)3 = 43
=> x + 1 = 4 => x = 3
b) x3 + 6x2 + 9x = 4x
=> x3 + 6x2 + 9x - 4x = 0
=> x3 + 6x2 + 5x = 0
=> x3 + 5x2 + x2 + 5x = 0
=> x2(x + 5) + x(x + 5) = 0
=> (x + 5)(x2 + x) = 0
=> (x + 5)x(x + 1) = 0
=> \(\hept{\begin{cases}x=-5\\x=0\\x=-1\end{cases}}\)
c) 4(x - 2)2 = (x + 2)2
=> 4(x2 - 4x + 4) = x2 + 4x + 4
=> 4x2 - 16x + 16 = x2 + 4x + 4
=> 4x2 - 16x + 16 - x2 - 4x - 4 = 0
=> 3x2 - 20x + 12 = 0
=> 3x2 - 18x - 2x + 12 = 0
=> 3x(x - 6) - 2(x - 6) = 0
=> (x - 6)(3x - 2) = 0
=> \(\orbr{\begin{cases}x=6\\x=\frac{2}{3}\end{cases}}\)
d) x4 - 16x2 = 0
=> x2(x2 - 16) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^2=16\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
e) x4 - 4x3 + x2 - 4x = 0
=> x4 + x2 - 4x3 - 4x = 0
=> x2(x2 + 1) - 4x(x2 + 1) = 0
=> (x2 - 4x)(x2 + 1) = 0
=> x(x - 4)(x2 + 1) = 0
=> \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)(vì x2 + 1 \(\ge\)1 > 0 \(\forall\)x)
f) x3 + x = 0 => x(x2 + 1) = 0 => x = 0 (vì x2 + 1 \(\ge1>0\forall\)x)
Bài 2 :
a, Ta có : \(\left(x+4\right)\left(x-1\right)=0\)
=> \(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)
b, Ta có : \(\left(3x-2\right)\left(4x-7\right)=0\)
=> \(\left[{}\begin{matrix}3x-2=0\\4x-7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2\\4x=7\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=\frac{7}{4}\end{matrix}\right.\)
c, Ta có : \(\left(x+5\right)\left(x^2+1\right)=0\)
=> \(\left[{}\begin{matrix}x+5=0\\x^2+1=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-5\\x^2+1=0\left(VL\right)\end{matrix}\right.\)
d, Ta có : \(x\left(x-1\right)\left(x^2+4\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\x-1=0\\x^2+4=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=0\\x=1\\x^2+4=0\left(VL\right)\end{matrix}\right.\)
e, Ta có : \(\left(3x+2\right)\left(x+\frac{1}{2}\right)=0\)
=> \(\left[{}\begin{matrix}3x+2=0\\x+\frac{1}{2}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{1}{2}\end{matrix}\right.\)
f, Ta có : \(\left(x+2\right)\left(x+3\right)\left(x^2+7\right)=0\)
=> \(\left[{}\begin{matrix}x+2=0\\x-3=0\\x^2+7=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-2\\x=3\\x^2+7=0\left(VL\right)\end{matrix}\right.\)
Bài 1 :
a, Ta có : \(1-\frac{x+3}{4}-\frac{x-2}{6}=0\)
=> \(\frac{12}{12}-\frac{3\left(x+3\right)}{12}-\frac{2\left(x-2\right)}{12}=0\)
=> \(12-3\left(x+3\right)-2\left(x-2\right)=0\)
=> \(12-3x-9-2x+4=0\)
=> \(-5x=-7\)
=> \(x=\frac{7}{5}\)
câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!
vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=\left(3x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)[\left(3x-2\right)-\left(x-1\right)]=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)
\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)
Chúc bạn học tốt!!
d/
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
e/
\(\Leftrightarrow x^3+x^2-6x-x^2-x+6=0\)
\(\Leftrightarrow x\left(x^2+x-6\right)-\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
Ta có
( 2 x 4 – 3 x 3 + x 2 ) : - 1 2 x 2 + 4 ( x – 1 ) 2 = 0 ⇔ 2 x 4 : ( - 1 2 x 2 ) - 3 x 3 : ( - 1 2 x 2 ) + x 2 : ( - 1 2 x 2 ) + 4 ( x 2 - 2 x + 1 ) = 0 ⇔ - 4 x 2 + 6 x – 2 + 4 x 2 – 8 x + 4 = 0
ó -2x + 2 = 0
ó x = 1
Đáp án cần chọn là: C
Câu c nha bạn