Tìm x
a, (x+7)\(^2\)-x(x-3)=12
b, x\(^2\)-3x+2=0
a) \(\sqrt{3x^2-5x+7}\)+\(\sqrt{3x^2+x+1}\) = 12x-12
b) \(\sqrt{x^2+33}\)+3 = 2x+\(\sqrt{x^2-12}\)
c) 3x-\(8\sqrt{x+14}\) = \(2\sqrt{2x-3}\) - 28
d) \(x^2\)+\(\sqrt{x+7}\) = 7
tìm x: [(3x^6)-(4x^3)]:x^3-(3x+1)^2:(3x+1)-3x^7:x^5=0
\(\dfrac{3x^6-4x^3}{x^3}-\dfrac{\left(3x+1\right)^2}{3x+1}-\dfrac{3x^7}{x^5}=0\)
\(\Leftrightarrow3x^3-4-3x-1-3x^2=0\)
\(\Leftrightarrow3x^3-3x^2-3x-5=0\)
\(\Leftrightarrow x\simeq1,9506\)
TÌM X
(2x-3)^2-4x-9=0
(3x-7).(3x+7)-54x-9x^2=0
(x+8)^2-x^2= 24
(x-2).(x^2+2x+4)+8-4x= 0
Em cảm ơn ạ
cho A=\(\left(\frac{2}{x^2-3x}-\frac{1}{x-3}\right)\cdot\frac{x^2-6x+9}{x-2}\)
a,Rút gọn A
b,tìm x để A>0
c,khi x>0,x khác 3 hãy tìm MinP=A+3x
BT1: Tìm x biết.
a, (5x-1)(2x+7)-(x+1)(6x-5)=16
b, (10x+9).x-(5x-1).(2x+3)=8
c, (3x-5).(7-5x)+(5x+2).(3x+2)-2=0
d, x.(x+1).(x+6)-x3=5x
a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42
1.Tìm x
a.x^2-5x-6=0
b.x^2+4x+11=0
c.|2x-5|+|4x-7|+3x-1=|2x-3|
d.(x-1)(x+3)-(x-7)(x+2)=x-5
a) x2 - 5x - 6 = 0
=> x2 - 2x - 3x - 6 = 0
=> (x2 - 2x) + (-3x - 6) = 0
=> x(x - 2) - 3 (x - 2) = 0
=> (x - 2) (x - 3) = 0
=> x - 2 = 0 => x = 2
x - 3 = 0 => x = 3
còn lại tương tự nhé!! 46566578768698945635655675656788787868789789879789098089364556546
Tìm x:
(2/3x-4/9) (1/2+ -3/7:x)=0
\(\left(\frac{2}{3}x-\frac{4}{9}\right)\left[\frac{1}{2}+\left(-\frac{3}{7}:x\right)\right]=0\)
\(\Rightarrow\left(\frac{2}{3}x-\frac{4}{9}\right)\left(\frac{1}{2}-\frac{3}{7}.\frac{1}{x}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}x-\frac{4}{9}=0\\\frac{1}{2}-\frac{3}{7}.\frac{1}{x}=0\end{cases}\Rightarrow\orbr{\begin{cases}\frac{2}{3}x=\frac{4}{9}\\\frac{1}{x}=\frac{7}{6}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}}\)
Vậy x = 2/3 , x = 6/7
Giải phương trình : a) (x^2+x+1)(x^2+x+2)=12
b) x^3+5x^2-10x-8=0
a) \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)
\(\Leftrightarrow\left(x^2+x+1\right)^2+\left(x^2+x+1\right)-12=0\)
\(\Leftrightarrow\left(x^2+x+1\right)^2-3\left(x^2+x+1\right)+4\left(x^2+x+1\right)-12=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x+1-3\right)+ 4\left(x^2+x+1-3\right)=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x+5\right)=0\)
\(\Leftrightarrow x^2+x+4=0\) hay \(x^2+x-2=0\)
\(\Leftrightarrow x^2+2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{15}{4}=0\) hay \(x^2-x+2x-2=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) (pt vô nghiệm) hay\(x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow x=1\) hay \(x=-2\)
-Vậy \(S=\left\{1;-2\right\}\)
b) \(x^3+5x^2-10x-8=0\)
\(\Leftrightarrow x^3-2x^2+7x^2-14x+4x-8=0\)
\(\Leftrightarrow x^2\left(x-2\right)+7x\left(x-2\right)+4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+7x+4\right)=0\)
\(\Leftrightarrow x=2\) hay \(x^2+2.\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{33}{4}=0\)
\(\Leftrightarrow x=2\) hay \(\left(x+\dfrac{7}{2}\right)^2-\dfrac{33}{4}=0\)
\(\Leftrightarrow x=2\) hay \(\left(x+\dfrac{7}{2}+\dfrac{\sqrt{33}}{2}\right)\left(x+\dfrac{7}{2}-\dfrac{\sqrt{33}}{2}\right)=0\)
\(\Leftrightarrow x=2\) hay \(x=\dfrac{-7-\sqrt{33}}{2}\) hay \(x=\dfrac{-7+\sqrt{33}}{2}\)
-Vậy \(S=\left\{2;\dfrac{-7-\sqrt{33}}{2};\dfrac{-7+\sqrt{33}}{2}\right\}\)
giải các bất phương trình sau
a, 3x-5 ≥ 2(x-6) -12
b, 2 (5-2x) ≥ 3-x
c, 2 ( -2x+1) ≤ -x+3
d, 2( x+1) ≤ -x+3
a: Ta có: \(3x-5\ge2\left(x-6\right)-12\)
\(\Leftrightarrow3x-5\ge2x-24\)
hay \(x\ge-19\)
b: Ta có: \(2\left(5-2x\right)\ge3-x\)
\(\Leftrightarrow10-4x-3+x\ge0\)
\(\Leftrightarrow-3x\ge-7\)
hay \(x\le\dfrac{7}{3}\)