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Lời giải:
a. PT $\Leftrightarrow (3-2x-3-2x)(3-2x+3+2x)=8$
$\Leftrightarrow -4x.6=8$
$\Leftrightarrow -24x=8\Leftrightarrow x=\frac{-1}{3}$
b.
$9x^5-72x^2=0$
$\Leftrightarrow 9x^2(x^3-8)=0$
$\Leftrightarrow x^2=0$ hoặc $x^3=8$
$\Leftrightarrow x=0$ hoặc $x=2$
c.
$5x^4-8x^2-4=0$
$\Leftrightarrow 5x^4-10x^2+2x^2-4=0$
$\Leftrightarrow 5x^2(x^2-2)+2(x^2-2)=0$
$\Leftrightarrow (5x^2+2)(x^2-2)=0$
$\Leftrightarrow 5x^2+2=0$ (loại) hoặc $x^2-2=0$ (chọn)
$\Leftrightarrow x=\pm \sqrt{2}$
d.
PT $\Leftrightarrow [x^2(x+1)-4(x+1)]:(x-2)=0$
$\Leftrightarrow (x^2-4)(x+1):(x-2)=0$
$\Leftrightarrow (x-2)(x+2)(x+1):(x-2)=0$
$\Leftrightarrow (x+2)(x+1)=0$
$\Leftrightarrow x+2=0$ hoặc $x+1=0$
$\Leftrightarrow x=-2$ hoặc $x=-1$
a: Ta có: \(\left(3-2x\right)^2-\left(3+2x\right)^2=8\)
\(\Leftrightarrow9-12x+4x^2-9-12x-4x^2=8\)
\(\Leftrightarrow-24x=8\)
hay \(x=-\dfrac{1}{3}\)
b: Ta có: \(9x^5-72x^2=0\)
\(\Leftrightarrow9x^2\left(x^3-8\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)
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
a) x³y + x - y - 1
= (x³y - y) + (x - 1)
= y(x³ - 1) + (x - 1)
= y(x - 1)(x² + x + 1) + (x - 1)
= (x - 1)[y(x² + x + 1) + 1]
= (x - 1)(x²y + xy + y + 1)
b) x²(x - 2) + 4(2 - x)
= x²(x - 2) - 4(x - 2)
= (x - 2)(x² - 4)
= (x - 2)(x - 2)(x + 2)
= (x - 2)²(x + 2)
c) x³ - x² - 20x
= x(x² - x - 20)
= x(x² + 4x - 5x - 20)
= x[(x² + 4x) - (5x + 20)]
= x[x(x + 4) - 5(x + 4)]
= x(x + 4)(x - 5)
d) (x² + 1)² - (x + 1)²
= (x² + 1 - x - 1)(x² + 1 + x + 1)
= (x² - x)(x² + x + 2)
= x(x - 1)(x² + x + 2)
e) 6x² - 7x + 2
= 6x² - 3x - 4x + 2
= (6x² - 3x) - (4x - 2)
= 3x(2x - 1) - 2(2x - 1)
= (2x - 1)(3x - 2)
f) x⁴ + 8x² + 12
= x⁴ + 2x² + 6x² + 12
= (x⁴ + 2x²) + (6x² + 12)
= x²(x² + 2) + 6(x² + 2)
= (x² + 2)(x² + 6)
g) (x³ + x + 1)(x³ + x) - 2
Đặt u = x³ + x
x³ + x + 1 = u + 1
(u + 1).u - 2
= u² + u - 2
= u² - u + 2u - 2
= (u² - u) + (2u - 2)
= u(u - 1) + 2(u - 1)
= (u - 1)(u + 2)
= (x³ + x - 1)(x³ + x + 2)
= (x³ + x - 1)(x³ + x² - x² - x + 2x + 2)
= (x³ + x - 1)[(x³ + x²) - (x² + x) + (2x + 2)]
= (x³ + x - 1)[x²(x + 1) - x(x + 1) + 2(x + 1)]
= (x³ + x - 1)(x - 1)(x² - x + 2)
h) (x + 1)(x + 2)(x + 3)(x + 4) - 1
= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 1
= (x² + 5x + 4)(x² + 5x + 6) - 1 (1)
Đặt u = x² + 5x + 4
u + 2 = x² + 5x + 6
(1) u.(u + 2) - 1
= u² + 2u - 1
= u² + 2u + 1 - 2
= (u² + 2u + 1) - 2
= (u + 1)² - 2
= (u + 1 + √2)(u + 1 - √2)
= (x² + 5x + 4 + 1 + √2)(x² + 5x + 4 + 1 - √2)
= (x² + 5x + 5 + √2)(x² + 5x + 5 - √2)
1. th1: 2x-5= 2-x nếu 2x-5>0 => x>\(\dfrac{5}{2}\)
2x-5=2-x
3x= 7
x=\(\dfrac{7}{3}\)(loại)
th2: 2x-5= -2-x nếu 2x-5 <0 => x<\(\dfrac{5}{2}\)
2x-5= -2-x
x= -1 (chọn)
S={-1}
2.|3x-2|+x=11
|3x-2| = 11-x
th1: 3x-2=11-x nếu 3x-2 >0 => x>\(\dfrac{2}{3}\)
3x-2= 11-x
4x= 13
x=\(\dfrac{13}{4}\)(chọn)
th2: 3x-2= -11-x nếu 3x-2<0 => x< \(\dfrac{2}{3}\)
3x-2=-11-x
4x= -9
x= \(\dfrac{-9}{4}\)(chọn)
S={\(\dfrac{13}{4}\); \(\dfrac{-9}{4}\)}
3. th1: 2x-1= 1-2x nếu 2x-1>0 => x>\(\dfrac{1}{2}\)
2x-1=1-2x
4x= 2
x=\(\dfrac{1}{2}\) (chọn)
th2: 2x-1=-1-2x nếu 2x-1<0 => x<\(\dfrac{1}{2}\)
2x-1=-1-2x
4x= 0
x= 0 (chọn)
S={\(\dfrac{1}{2}\); 0}
Bài 2:
a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)
=>-13x=26
hay x=-2
b: \(\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{5}\right\}\)
c: \(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
hay \(x\in\left\{-5;2\right\}\)
Bài 1:
\(a,x^4+5x^2+9\\=(x^4+6x^2+9)-x^2\\=[(x^2)^2+2\cdot x^2\cdot3+3^2]-x^2\\=(x^2+3)^2-x^2\\=(x^2+3-x)(x^2+3+x)\)
\(b,x^4+3x^2+4\\=(x^4+4x^2+4)-x^2\\=[(x^2)^2+2\cdot x^2\cdot2+2^2]-x^2\\=(x^2+2)^2-x^2\\=(x^2+2-x)(x^2+2+x)\)
\(c,2x^4-x^2-1\\=2x^4-2x^2+x^2-1\\=2x^2(x^2-1)+(x^2-1)\\=(x^2-1)(2x^2+1)\\=(x-1)(x+1)(2x^2+1)\)
Bài 2:
\(a,\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=120\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x+4\right)\right]\cdot\left[\left(x+2\right)\left(x+3\right)\right]=120\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=120\) (1)
Đặt \(x^2+5x+5=y\), khi đó (1) trở thành:
\(\left(y-1\right)\left(y+1\right)=120\)
\(\Leftrightarrow y^2-1=120\)
\(\Leftrightarrow y^2=121\)
\(\Leftrightarrow\left[{}\begin{matrix}y=11\\y=-11\end{matrix}\right.\)
+, TH1: \(y=11\Leftrightarrow x^2+5x+5=11\)
\(\Leftrightarrow x^2+5x-6=0\)
\(\Leftrightarrow x^2-x+6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-6\end{matrix}\right.\left(\text{nhận}\right)\)
+, TH2: \(y=-11\Leftrightarrow x^2+5x+5=-11\)
\(\Leftrightarrow x^2+5x+16=0\)
\(\Leftrightarrow\left[x^2+2\cdot x\cdot\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2\right]-\dfrac{25}{4}+16=0\)
\(\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
Ta thấy: \(\left(x+\dfrac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}\ge\dfrac{39}{4}>0\forall x\)
Mà \(\left(x+\dfrac{5}{2}\right)^2+\dfrac{39}{4}=0\)
\(\Rightarrow\) loại
Vậy \(x\in\left\{1;-6\right\}\).
\(b,\) Đề thiếu vế phải rồi bạn.
d: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
\(\Leftrightarrow x\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
a)
\(-4x\left(-2x+1\right):-4x-\left(x+2\right)=8\)
\(-2x+1-x-2=8\)
\(-3x-1=8\)
\(-3x=9\)
\(x=-3\)
b)
\(-\frac{1}{2}x^2\left(-4x^2+6x-2\right):\left(\frac{-1}{2}x^2\right)+4\left(x^2-2x+1\right)==0\)
\(-4x^2+6x-2+4x^2-8x+4=0\)
\(-2x+2=0\)
\(-2x=-2\)
\(x=1\)
ggaao ko