TÌM X
a) \(3x^2-2x-5=0\)
b)\(x^2-x-6=0\)
c) \(x^2-12x+35=0\)
Đề: Tìm x
a)2x.(3x+5)-x.(6x-1)=33 k)5(x+3)-2x(x+3)=0
b)x(3x-1)+12x-4=0 i)5x(x-2)-(2-x)=0
c)5x(2x+1)-12x-6=0 m)x(x-1)-2(1-x)=0
d)x3-5x2+4x-20=0
e)2x3-5x2+2x-5=0
g)(x-2)3-x(x+1).(x-1)+62=5
a)2x.(3x+5)-x.(6x-1)=33
=>\(6x^2+10x-6x^2+x=33\)
=>11x=33
=>x=3
b)x(3x-1)+12x-4=0
=>x(3x-1)+4(3x-1)=0
=>(x-4)(3x-1)=0
=>x-4=0 hoặc 3x-1=0
+)x-4=0 +)3x-1=0
=>x=4 =>x=\(\frac{1}{3}\)
c)5x(2x+1)-12x-6=0
=>10x\(^2\)+5x-12x-6=0
=>10x\(^2\)-7x-6=0
=>(10x\(^2\)+5x)-(12x+6)=0
=>5x(2x+1)-6(2x+1)=0
=>(5x-6)(2x+1)=0
=>\(\left[{}\begin{matrix}5x-6=0\\2x+1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\frac{6}{5}\\x=\frac{-1}{2}\end{matrix}\right.\)
Giai phường trình sau:
a, \(3x^2+2x-1=0\) e, \(4x^2-12x+5=0\) i,\(2x^2+5x-3=0\)
b,\(x^2-5x+6=0\) f, \(2x^2+5x+3=0\) j,\(x^2+6x-16=0\)
c,\(x^2-3x+2=0\) g,\(x^2+x-2=0\)
d,\(2x^2-6x+1=0\) h, \(x^2-4x+3=0\)
a) Ta có: \(3x^2+2x-1=0\)
\(\Leftrightarrow3x^2+3x-x-1=0\)
\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)
b) Ta có: \(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: S={2;3}
c) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: S={1;2}
d) Ta có: \(2x^2-6x+1=0\)
\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)
mà \(2\ne0\)
nên \(x^2-3x+\dfrac{1}{3}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)
e) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)
Dùng công thức nghiệm,giải các phương trình sau:
a. \(x^2+3x+4=0\)
b. \(4x^2-4x+1=0\)
c. \(x^2-5x-6=0\)
d. \(3x^2+12x-2=0\)
e. \(x^2+2\sqrt{5}x-1=0\)
f. \(2x^2-4\sqrt{2}x+2=0\)
Bài 4: Tìm x, biết.
a) 4x(x - 7) - 4x2 = 56
b) 12x(3x - 2) - (4 - 6x) = 0
c) 4(x - 5) - (5 - x)2 = 0
d) x(x +1) - x(x - 3) = 0
e) - 6x + 8 = 0 f) 2 + 2x + = 0
c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
I) THỰC HIỆN PHÉP TÍNH a) 2x(x^2-4y) b)3x^2(x+3y) c) -1/2x^2(x-3) d) (x+6)(2x-7)+x e) (x-5)(2x+3)+x II phân tích đa thức thành nhân tử a) 6x^2+3xy b) 8x^2-10xy c) 3x(x-1)-y(1-x) d) x^2-2xy+y^2-64 e) 2x^2+3x-5 f) 16x-5x^2-3 g) x^2-5x-6 IIITÌM X BIẾT a)2x+1=0 b) -3x-5=0 c) -6x+7=0 d)(x+6)(2x+1)=0 e)2x^2+7x+3=0 f) (2x-3)(2x+1)=0 g) 2x(x-5)-x(3+2x)=26 h) 5x(x-1)=x-1 IV TÌM GTNN,GTLN. a) tìm giá trị nhỏ nhất x^2-6x+10 2x^2-6x b) tìm giá trị lớn nhất 4x-x^2-5 4x-x^2+3
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
Giải các phương trình tích sau:
1.a)(3x – 2)(4x + 5) = 0 b) (2,3x – 6,9)(0,1x + 2) = 0
c)(4x + 2)(x2 + 1) = 0 d) (2x + 7)(x – 5)(5x + 1) = 0
2. a)(3x + 2)(x2 – 1) = (9x2 – 4)(x + 1)
b)x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0
c)2x(x – 3) + 5(x – 3) = 0 d)(3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
3.a)(2x – 5)2 – (x + 2)2 = 0 b)(3x2 + 10x – 8)2 = (5x2 – 2x + 10)2
c)(x2 – 2x + 1) – 4 = 0 d)4x2 + 4x + 1 = x2
4. a) 3x2 + 2x – 1 = 0 b) x2 – 5x + 6 = 0
c) x2 – 3x + 2 = 0 d) 2x2 – 6x + 1 = 0
e) 4x2 – 12x + 5 = 0 f) 2x2 + 5x + 3 = 0
Bài 1:
a) (3x - 2)(4x + 5) = 0
<=> 3x - 2 = 0 hoặc 4x + 5 = 0
<=> 3x = 2 hoặc 4x = -5
<=> x = 2/3 hoặc x = -5/4
b) (2,3x - 6,9)(0,1x + 2) = 0
<=> 2,3x - 6,9 = 0 hoặc 0,1x + 2 = 0
<=> 2,3x = 6,9 hoặc 0,1x = -2
<=> x = 3 hoặc x = -20
c) (4x + 2)(x^2 + 1) = 0
<=> 4x + 2 = 0 hoặc x^2 + 1 # 0
<=> 4x = -2
<=> x = -2/4 = -1/2
d) (2x + 7)(x - 5)(5x + 1) = 0
<=> 2x + 7 = 0 hoặc x - 5 = 0 hoặc 5x + 1 = 0
<=> 2x = -7 hoặc x = 5 hoặc 5x = -1
<=> x = -7/2 hoặc x = 5 hoặc x = -1/5
bài 2:
a, (3x+2)(x^2-1)=(9x^2-4)(x+1)
(3x+2)(x-1)(x+1)=(3x-2)(3x+2)(x+1)
(3x+2)(x-1)(x+1)-(3x-2)(3x+2)(x+1)=0
(3x+2)(x+1)(1-2x)=0
b, x(x+3)(x-3)-(x-2)(x^2-2x+4)=0
x(x^2-9)-(x^3+8)=0
x^3-9x-x^3-8=0
-9x-8=0
tự tìm x nha
a, (3x-2) (x+6) (x^2 +5) = 0
b, (2x+5)^2 = (3x-1)^2
c, 4x^2 (x-1) - x+1 = 0
d, 9 (2x+1) = 4(x-5)^2
e, x^3 - 4x^2 - 12x +27 = 0
f, x^3 + 3x^2 -6x -8 =0
đề là gì
a)\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}3x-2=0\\x+6=0\\x^2+5=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2\\x=-6\\x^2=-5\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{2}{3}\\x=-6\\x\in\varnothing\end{cases}}}\)
vậy x=2/3 hoặc x=-6
a, (3x-2) (x+6) (x^2 +5) = 0
<=> 3x - 2 = 0 hoặc x + 6 = 0 hoặc x2 + 5 = 0 (loại vì x2 \(\ge\)0 => x2 + 5 > 0)
<=> x = 2/3 hoặc x = -6
b, (2x+5)^2 = (3x-1)^2
<=> (2x + 5)2 - (3x - 1)2 = 0
<=> (2x + 5 - 3x + 1)(2x + 5 + 3x - 1) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-3x+6=0\\2x+3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}-x=-6\\5x=4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=6\\x=\frac{4}{5}\end{cases}}}\)
c, 4x2 (x-1) - x+1 = 0
<=> 4x2(x - 1) - (x - 1) = 0
<=> (x - 1)(4x2 - 1) = 0
<=> (x - 1)(2x - 1)(2x + 1) = 0
vậy x - 1 = 0 hoặc 2x - 1 = 0 hoặc 2x + 1 = 0
hay x = 1 hoặc x = 1/2 hoặc x = -1/2
a) 5x+3=2x-8
b) 6x-3(x+2)=5x+3
c) (3x-9)(5x+10)=0
d) 8x(x+2)+16(x+2)=0
e) x2 -12x+35=0
a) 5x +3=2x-8 <=>5x-2x=-8-3 <=>3x=-11 <=> x=\(\dfrac{-11}{3}\)
b)6x-3(x+2)=5x+3<=> (6-3-5)x-9=0 <=> x=\(\dfrac{-9}{2}\)
c) (3x-9)(5x+10)=0<=> \(\left[{}\begin{matrix}3x-9=0\\5x+10=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
d)8x(x+2)+16(x+2)=0<=>(x+2)(8x+16)=0<=>\(\left[{}\begin{matrix}x=-2\\x=-2\end{matrix}\right.\)
e)x2 -12x+35=0 <=>\(\left[{}\begin{matrix}x=7\\x=5\end{matrix}\right.\)
Bài 8: Giải các phương trình tích sau:
4. a) 3x2 + 2x – 1 = 0 b) x2 – 5x + 6 = 0
c) x2 – 3x + 2 = 0 d) 2x2 – 6x + 1 = 0
e) 4x2 – 12x + 5 = 0 f) 2x2 + 5x + 3 = 0
g) x2 + x – 2 = 0 h) x2 – 4x + 3 = 0
i) 2x2 + 5x – 3 = 0 j) x2 + 6x – 16 = 0
5. a) 3x2 + 12x – 66 = 0 b) 9x2 – 30x + 225 = 0
c) x2 + 3x – 10 = 0 d) 3x2 – 7x + 1 = 0
e) 3x2 – 7x + 8 = 0 f) 4x2 – 12x + 9 = 0
g) 3x2 + 7x + 2 = 0 h) x2 – 4x + 1 = 0
i) 2x2 – 6x + 1 = 0 j) 3x2 + 4x – 4 = 0