a) 36x2-25-x(6x-5)=0
b) 1 - x+1/2 = 2x-1/4 -1/2
c) 1 - x/x+1 = 7x/(3x-2)(x+1) + 2/2-3x
Bài 1: Giải các phương trình sau:
a) 3x ^ 2 - 5x + 2 = 0
d) - 4x ^ 2 + 25 = 0
b) 11x - 2x ^ 2 = 0
e) sqrt(x ^ 2 - x + 9) = 2x + 1
c) x ^ 2 + 5x + 7 = 0
f) 6x ^ 4 - 7x ^ 2 + 1 = 0
a: =>3x^2-3x-2x+2=0
=>(x-1)(3x-2)=0
=>x=2/3 hoặc x=1
b: =>2x^2=11
=>x^2=11/2
=>\(x=\pm\dfrac{\sqrt{22}}{2}\)
c: Δ=5^2-4*1*7=25-28=-3<0
=>PTVN
f: =>6x^4-6x^2-x^2+1=0
=>(x^2-1)(6x^2-1)=0
=>x^2=1 hoặc x^2=1/6
=>\(\left[{}\begin{matrix}x=\pm1\\x=\pm\dfrac{\sqrt{6}}{6}\end{matrix}\right.\)
d: =>(5-2x)(5+2x)=0
=>x=5/2 hoặc x=-5/2
e: =>4x^2+4x+1=x^2-x+9 và x>=-1/2
=>3x^2+5x-8=0 và x>=-1/2
=>3x^2+8x-3x-8=0 và x>=-1/2
=>(3x+8)(x-1)=0 và x>=-1/2
=>x=1
Bài 1:
a) 7x –12 = 5x + 3
b) 2(3x –5) –7(x + 1) = 2
c) (1 –3x)^2= (4x –3)^2
d) (2x + 3)(4x –2) –2(2x + 1)^2= 12
Bài 2:
Cho biểu thứcA = (5x –3y + 1)(7x + 2y –2)
a) Tìm x sao cho với y = 2 thì A = 0
b) Tìm y sao cho với x = -2 thì A = 0
1.
a.\(\Leftrightarrow7x-5x=3+12\)
\(\Leftrightarrow2x=15\Leftrightarrow x=\dfrac{15}{2}\)
b.\(\Leftrightarrow6x-10-7x-7=2\)
\(\Leftrightarrow x=-19\)
c.\(\Leftrightarrow1-3x=4x-3\)
\(\Leftrightarrow7x=2\Leftrightarrow x=\dfrac{2}{7}\)
d.\(\Leftrightarrow8x^2-4x+12x-6-8x^2-8x-2=12\)
\(\Leftrightarrow-2=12\left(voli\right)\)
Bài 3: Giải các phương trình sau:
a, 2x3 - 50x = 0
b, 2x (3x - 5) - (5 - 3x)
c, 9(3x - 2) = x(2 - 3x)
d, (2x - 1)2 - 25 = 0
e, 25x2 - 2 = 0
f, x2 - 25 = 6x - 9
g, 5x(x - 3) - 2x + 6 = 0
h, 3x(x - 7) - 2(x - 7) = 0
i, 7x2 - 28 = 0
j, (2x + 1) + x(2x + 1) = 0
k, (x + 2)2 - (x - 2)(x + 2) = 0
l, x3 + 5x2 - 4x - 20 = 0
m, x2 - 25 + 2(x + 5) = 0
n, x3 - 3x + 2 = 0
o, x2 - 6x + 8 = 0
p, x2 - 5x - 14 = 0
q, (x - 2)2 - (x - 3)(x + 3) = 6
r, (2x - 1)2 - (2x + 5)(2x - 5) = 18
Bài 7. Tìm x,biết:
a) x-3x2=0 e) 5x(3x-1)+x(3x-1)-2(3x-1)=0
b) (x+3)2-x(x-2)=13 c) (x-4)2-36=0
d) x2-7x+12=0 g) x2-2018x-2019=0
Bài 8. Tìm x, biết
a) (2x-1)2=(x+5)2 b) x2-x+1/4
c) 4x4-101x2+25=0 d) x3-3x2+9x-91=0
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)
Bài 1:Tìm min hoặc max của biểu thức:
a. x^2-6x+15
b.3x^2-15x-4
c.7x-2x^2
Bài 2: tìm x:
a. x^2-25+2(x+5)=0
b. 2(x^2+8x+16)-x^2+4=0
c. x^2(x-2)+7x=14
giúp mik vs sắp phải nộp rồi TvT
Bài 1:
a) \(x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\)
b) \(3x^2-15x+4=3\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{59}{4}=3\left(x-\dfrac{5}{2}\right)^2-\dfrac{59}{4}\ge-\dfrac{59}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)
Bài 2:
a) \(\Rightarrow\left(x-5\right)\left(x+5\right)+2\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)
c) \(\Rightarrow x^2\left(x-2\right)+7\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x^2+7\right)=0\)
\(\Rightarrow x=2\left(do.x^2+7\ge7>0\right)\)
Bài 2: giải phương trình sau
a) \(X^4\)-\(x^2\)-2=0
b) (x+1)\(^4\)-x\(^2\)+2)\(^2\)=0
c)3x\(^2\)-2x-8=0
Bài 3: giải phương trình sau
a) x\(^3\)-0,25=0
b) x\(^4\)+2x\(^3\)+x\(^2\)=0
c) x\(^3\)-1=0
d) 6x\(^2\)-7x+2=0
Mong có người giải giùm xin kẻm ơn :>
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
a) 2x2 + 2x(5 - x)=12 d) 2(x + 5) - x2 - 5x = 0 g) (3x + 1)2 - (x+1) = 0
b) (5 - 2x)2 - 16 = 0 e) (2x - 1)2 - 4(x + 7)(x - 7) = 0 h) x2 + 7x - 8 = 0
c) 3x2 - 3x(x-2) = 36 f) (x + 4)2 - (x + 1)(x - 1) = 16 i) -2x2 +13x -15 = 0
mik cần gấp, cảm ơn mọi người.
\(a,\Leftrightarrow2x^2+10x-2x^2=12\Leftrightarrow x=\dfrac{12}{10}=\dfrac{6}{5}\\ b,\Leftrightarrow\left(5-2x-4\right)\left(5-2x+4\right)=0\\ \Leftrightarrow\left(1-2x\right)\left(9-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{9}{2}\end{matrix}\right.\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\\ d,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ e,\Leftrightarrow4x^2-4x+1-4x^2+196=0\\ \Leftrightarrow-4x=-197\Leftrightarrow x=\dfrac{197}{4}\)
\(f,\Leftrightarrow x^2+8x+16-x^2+1=16\Leftrightarrow8x=-1\Leftrightarrow x=-\dfrac{1}{8}\\ g,Sửa:\left(3x+1\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(3x+1-x-1\right)\left(3x+1+x+1\right)=0\\ \Leftrightarrow2x\left(4x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\\ h,\Leftrightarrow x^2+8x-x-8=0\\ \Leftrightarrow\left(x+8\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-8\end{matrix}\right.\\ i,\Leftrightarrow2x^2-13x+15=0\\ \Leftrightarrow2x^2+2x-15x-15=0\\ \Leftrightarrow\left(x+1\right)\left(2x-15\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{15}{2}\end{matrix}\right.\)
giải các phương trình sau
a) (3x+1)2-6(2x-7)(x-3)=0
b) (3x+1)(x-3)2=(3x+1)(2x-5)2
c) 0,75x(x+5)=(x+5)(3-1,25x)
a: =>9x^2+6x+1-6(2x^2-13x+21)=0
=>9x^2+6x+1-12x^2+78x-126=0
=>-3x^2+84x-125=0
=>\(x\in\left\{26.42;1.58\right\}\)
b: =>(3x+1)[(2x-5)^2-(x-3)^2]=0
=>(3x+1)(2x-5-x+3)(2x-5+x-3)=0
=>(3x+1)(x-2)(3x-8)=0
=>\(x\in\left\{-\dfrac{1}{3};2;\dfrac{8}{3}\right\}\)
c; =>(x+5)(0,75x-3+1,25x)=0
=>(x+5)(2x-3)=0
=>x=3/2 hoặc x=-5