Giải các phương trình sau :
a,\(\left(x+3\right)^2+2\left(x-1\right)^2=\left(3x-7\right)\left(x-2\right)\)
b, \(\left(x-3\right)^2-3x=\left(x+2\right)^3+1\)
Giải các phương trình sau :
a,\(\left(x+3\right)^2+2\left(x-1\right)^2=\left(3x-7\right)\left(x-2\right)\)
b, \(\left(x-3\right)^2-3x=\left(x+2\right)^3+1\)
Giải các bất phương trình sau :
\(a.4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(b.\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
c. \(\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
Giải các phương trình
a) \(\left|x-2\right|\)=\(\left|x+3\right|\)
b) \(\left|3x+7\right|\)=\(\left|x-2\right|\)
c) \(\left|5-2x\right|\)=\(\left|3x-4\right|\)
a: =>x+3=x-2 hoặc x+3=2-x
=>2x=-1
=>x=-1/2
b: =>3x+7=x-2 hoặc 3x+7=-x+2
=>2x=-9 hoặc 4x=-5
=>x=-5/4 hoặc x=-9/2
c: =>|3x-4|=|2x-5|
=>3x-4=2x-5 hoặc 3x-4=-2x+5
=>x=-1 hoặc x=9/5
Giải các phương trình sau:
f. 5 – (x – 6) = 4(3 – 2x)
g. 7 – (2x + 4) = – (x + 4)
h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
i. \(\left(x-2^3\right)+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. 2x(x+2)\(^2\)−8x\(^2\)=2(x−2)(x\(^2\)+2x+4)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x=2x^3-16\)
<=>\(8x=-16\)
<=>\(x=-2\)
i. (x−2\(^3\))+(3x−1)(3x+1)=(x+1)\(^3\)
<=>\(x-8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(6x^2-2x-10=0\)
<=>\(3x^2-x-5=0\)
<=>\(\left[{}\begin{matrix}x=\dfrac{1+\sqrt{61}}{6}\\x=\dfrac{1-\sqrt{61}}{6}\end{matrix}\right.\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>\(x=\dfrac{1}{5}\)
f. 5 – (x – 6) = 4(3 – 2x)
<=>5-x+6=12-8x
<=>7x=1
<=>x=\(\dfrac{1}{7}\)
g. 7 – (2x + 4) = – (x + 4)
<=>7-2x-4=-x-4
<=>x=7
h. \(2x\left(x+2\right)^2-8x^2=2\left(x-2\right)\left(x^2+2x+4\right)\)
<=>\(2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
<=>\(2x^3+8x^2+8x-8x^2=2x^3-16\)
<=>\(8x=-16\)
<=>x=-2
i.\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)
<=>\(x^3-6x^2+12x+8+9x^2-1=x^3+3x^2+3x+1\)
<=>\(9x+6=0\)
<=>x=\(\dfrac{-2}{3}\)
k. (x + 1)(2x – 3) = (2x – 1)(x + 5)
<=>\(2x^2-x-3=2x^2+9x-5\)
<=>10x=2
<=>
Bài tập. Giải các phương trình sau:
a) \(\left|7-x\right|+2x=3\)
b) \(\left|2x-3\right|-4x-9=0\)
c) \(\left|3x+5\right|=\left|2-5x\right|\)
d) \(x\left|x-3\right|-\left|x^2+x+1\right|=1\)
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
Giải các phương trình:
\(a,\left(x+3\right)^2+2\left(x-1\right)^2=\left(3x-7\right)\left(x-2\right)\)
\(b,\left(x-4\right)\left(x-3\right)=\left(x-4\right)^2\)
a, (x+3)^2 + 2(x-1)^2 = (3x-7)(x-2)
<=> x^2 + 6x + 9 + 2x^2 - 4x + 2 = 3x^2 - 13x + 14
<=> 15x - 3 = 0
<=> x = 1/5
Vậy x=1/5 là nghiệm của phương trình
b, ( x - 4)( x - 3)= (x-4)^2
Đặt x - 4 = y ta có phương trình :
y(y +1 ) = y^2
<=> y^2+y= y^2
<=> y=0
=> x- 4 =0
<=> x=4
Vậy x=4 là nghiệm của phương trình
bài 1. giải các phương trình sau
a / \(x =(4x+1) (\frac{3x+7}{3-5x}+1)=(x+4)(\frac{3x+7}{5x-3}-1)\)
b/ \(\left(x^2+3x+1\right)\left(\frac{4x-3}{3x+1}+2\right)=\left(4x+7\right)\left(\frac{4x-3}{3x+1}+2\right)\)1)
bài 2. giải phương trình sau bằng cách đưa về phương trình tích
a/\(\left(4x-5\right)^2-2\left(16x^2-25\right)=0\)
b/ \(\left(4x+3\right)^2=4\left(x^2-2x+1\right)\)
c. \(3x^3-3x^2-6x=0\)
cảm ơn mọi người nhiều lắm !
cho hàm số \(f\left(x\right)=x^3-3x^2+2\)
a, giải bất phương trình \(f'\left(x\right)\le0\)
b, giải phương trình \(f'=\left(x^2-3x+2\right)=0\)
c, đặt \(g\left(x\right)=f\left(1-2x\right)+x^2-x+2022\) giải bất phương trình\(g'\left(x\right)\ge0\)
\(a,f'\left(x\right)=3x^2-6x\\ f'\left(x\right)\le0\Leftrightarrow3x^2-6x\le0\\ \Leftrightarrow3x\left(x-2\right)\le0\Leftrightarrow0\le x\le2\)
Lời giải:
a. $f'(x)\leq 0$
$\Leftrightarrow 3x^2-6x\leq 0$
$\Leftrightarrow x(x-2)\leq 0$
$\Leftrightarrow 0\leq x\leq 2$
b.
$f'(x)=x^2-3x+2=0$
$\Leftrightarrow 3x^2-6x=x^2-3x+2=0$
$\Leftrightarrow 3x(x-2)=(x-1)(x-2)=0$
$\Leftrightarrow x-2=0$
$\Leftrightarrow x=2$
c.
$g(x)=f(1-2x)+x^2-x+2022$
$g'(x)=(1-2x)'f(1-2x)'_{1-2x}+2x-1$
$=-2[3(1-2x)^2-6(1-2x)]+2x-1$
$=-24x^2+2x+5$
$g'(x)\geq 0$
$\Leftrightarrow -24x^2+2x+5\geq 0$
$\Leftrightarrow (5-12x)(2x-1)\geq 0$
$\Leftrightarrow \frac{-5}{12}\leq x\leq \frac{1}{2}$
Giải các phương trình sau
\(a.\left(3x+2\right)^2-\left(3x-2\right)^2=5x+8\)
\(b.3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(c.\left(x+3\right)^2-\left(x-3\right)^2=6x+18\)
\(d.\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x+2\right)\)
a) \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+8\)
\(\Rightarrow\left(3x+2+3x-2\right)\left(3x+2-3x+2\right)=5x+8\)
\(\Rightarrow4.6x=5x+8\Rightarrow24x=5x+8\)
\(\Rightarrow19x=8\Rightarrow x=\frac{8}{19}\)
b) \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Rightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)
\(\Rightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
\(\Rightarrow-12x+12+9x-9=3x-9\)
\(\Rightarrow-3x+3=3x-9\)
\(\Rightarrow6x=12\Rightarrow x=2\)
c) \(\left(x+3\right)^2-\left(x-3\right)^2=6x+18\)
\(\Rightarrow\left(x+3+x-3\right)\left(x+3-x+3\right)=6x+18\)
\(\Rightarrow2x.6=6x+18\Rightarrow6x=18\Rightarrow x=3\)
Giải các phương trình sau:
a) \(\left(4x-1\right)\left(x-3\right)=\left(x-3\right)\left(5x+2\right)\).
b)\(\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\)
c)\(\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\). Giải chi tiết hộ mik nhoa, mik tik
TA CÓ:
\(a,\left(4x-1\right)\left(x-3\right)=\left(x-3\right)\left(5x+2\right)\Leftrightarrow\left(4x-1\right)\left(x-3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\left(x-3\right)\left(4x-1-5x-2\right)=0\Leftrightarrow\left(x-3\right)\left(-x-3\right)=0\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
\(b,\left(x+3\right)\left(x-5\right)+\left(x+3\right)\left(3x-4\right)=0\Leftrightarrow\left(x+3\right)\left(x-5+3x-4\right)=0\)
\(\left(x-3\right)\left(4x-9\right)=0\orbr{\begin{cases}x=3\\x=\frac{9}{4}\end{cases}}\)
\(c,\left(1-x\right)\left(5x+3\right)=\left(3x-7\right)\left(x-1\right)\Leftrightarrow\left(1-x\right)\left(5x+3\right)=\left(7-3x\right)\left(1-x\right)\)
\(\left(1-x\right)\left(5x+3-7+3x\right)=0\Leftrightarrow\left(1-x\right)\left(8x-4\right)=0\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)