Tìm x biết rằng
a)x^3-4x^2+4x=0
b)x^3-9x=0
Những câu hỏi liên quan
tìm X
a) x^3-5x^2-9x=0
b)(4x-3)^2-3x(3-4x)=0
tìm x biết
1, x mũ 3 + 4x mũ 2 + 4x = 0
2, ( x + 3 ) mũ 2 - 4 = 0
3, x mũ 4 - 9x mũ 2 = 0
4, x mũ 2 - 6x + 9 = 81
5, x mũ 3 + 6x mũ 2 + 9x - 4x = 0
1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\Leftrightarrow x=-2;x=0\)
2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)
3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)
4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)
\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)
5, em xem lại đề nhé
à lag tý @@
5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)
\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)
a)\(x^3+4x^2+4x=0\)
\(\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)
b)\(\left(x+3\right)^2-4=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3-2=0\\x+3+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}}\)
c)\(x^4-9x^2=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)
d)\(x^2-6x+9=81\)
\(\Leftrightarrow\left(x-3\right)^2=81\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=9\\x-3=-9\end{cases}\Leftrightarrow\orbr{\begin{cases}x=12\\x=-6\end{cases}}}\)
e)\(x^3+6x^2+9x-4x=0\)
\(\Leftrightarrow x^3+6x^2+5x=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+5x=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0;x=-5\\x=-1\end{cases}}}\)
#H
a,x^2-9x+20=0
b,x^3-4x^2+5x=0
c,x^2=2x-15=0
d,(x^2-1)^2=4x+1
e,4x^3-9x^2+6x-1=0
f,x^4-4x^3-x^2+16x-12=0
a) Ta có: \(x^2-9x+20=0\)
\(\Leftrightarrow x^2-5x-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
Vậy: x∈{4;5}
b) Ta có: \(x^3-4x^2+5x=0\)
\(\Leftrightarrow x\left(x^2-4x+5\right)=0\)(1)
Ta có: \(x^2-4x+5\)
\(=x^2-4x+4+1=\left(x-2\right)^2+1\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)
hay \(x^2-4x+5>0\forall x\)(2)
Từ (1) và (2) suy ra x=0
Vậy: x=0
c) Sửa đề: \(x^2-2x-15=0\)
Ta có: \(x^2-2x-15=0\)
\(\Leftrightarrow x^2+3x-5x-15=0\)
\(\Leftrightarrow x\left(x+3\right)-5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
Vậy: x∈{-3;5}
d) Ta có: \(\left(x^2-1\right)^2=4x+1\)
\(\Leftrightarrow x^4-2x^2+1-4x-1=0\)
\(\Leftrightarrow x^4-2x^2-4x=0\)
\(\Leftrightarrow x\left(x^3-2x-4\right)=0\)
\(\Leftrightarrow x\left(x^3+2x^2+2x-2x^2-4x-4\right)=0\)
\(\Leftrightarrow x\cdot\left[x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\right]=0\)
\(\Leftrightarrow x\cdot\left(x^2+2x+2\right)\cdot\left(x-2\right)=0\)(3)
Ta có: \(x^2+2x+2\)
\(=x^2+2x+1+1=\left(x+1\right)^2+1\)
Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)
hay \(x^2+2x+2>0\forall x\)(4)
Từ (3) và (4) suy ra
\(\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy: x∈{0;2}
Tìm x biết:
a) \(3x^2-4x=0\). b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\).
c) \(9x^2+6x+1=0\). d) \(x^2-4x=4\).
a)\(3x^2-4x=0<=>x(3x-4)=0\)
TH1: x=0
TH2 3x-4=0 <=>x=4/3
KL:.....
b) (x+3)(x−1)+2x(x+3)=0.
<=> (x+3)(x-1+2x)=0
TH1: x+3=0 <=> x=-3
TH2 x-1=0 <=> x=1
KL:.....
c) \(9x^2+6x+1=0. <=>(3x+1)^2=0<=>3x+1=0<=>x=-1/3 \)
KL:......
d) \(x^2−4x=4.<=>(x-2)^2=0<=>x-2=0<=>x=2\)
KL:....
Đúng 0
Bình luận (0)
a) \(3x^2-4x=0\)
\(\Leftrightarrow x\left(3x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{3}\end{matrix}\right.\)
b) \(\left(x+3\right)\left(x-1\right)+2x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(9x^2+6x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^2=0\)
\(\Leftrightarrow3x+1=0\Leftrightarrow x=-\dfrac{1}{3}\)
d) \(x^2-4x=4\)
\(\Leftrightarrow\left(x-2\right)^2=8\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\sqrt{2}\\x-2=-2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}+2\\x=-2\sqrt{2}+2\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Tìm x biết.
a) 4x^2 - 49 = 0 b) x^2 + 36 = 12x
c) 1/16x^2 - x + 4 = 0 d) x^3 -3√3x2 + 9x - 3√3 = 0
e) (x - 2)^2 - 16 = 0 f) x^2 - 5x - 14 = 0
g) 8x(x - 3) + x - 3 = 0
a, 4x2 - 49 = 0
⇔⇔ (2x)2 - 72 = 0
⇔⇔ (2x - 7)(2x + 7) = 0
⇔{2x−7=02x+7=0⇔⎧⎪ ⎪⎨⎪ ⎪⎩x=72x=−72⇔{2x−7=02x+7=0⇔{x=72x=−72
b, x2 + 36 = 12x
⇔⇔ x2 + 36 - 12x = 0
⇔⇔ x2 - 2.x.6 + 62 = 0
⇔⇔ (x - 6)2 = 0
⇔⇔ x = 6
e, (x - 2)2 - 16 = 0
⇔⇔ (x - 2)2 - 42 = 0
⇔⇔ (x - 2 - 4)(x - 2 + 4) = 0
⇔⇔ (x - 6)(x + 2) = 0
⇔{x−6=0x+2=0⇔{x=6x=−2⇔{x−6=0x+2=0⇔{x=6x=−2
f, x2 - 5x -14 = 0
⇔⇔ x2 + 2x - 7x -14 = 0
⇔⇔ x(x + 2) - 7(x + 2) = 0
⇔⇔ (x + 2)(x - 7) = 0
⇔{x+2=0x−7=0⇔{x=−2x=7
a,\(4x^2-49=0\)
\(\Leftrightarrow\left(2x\right)^2-7^2=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{7}{2}\end{cases}}}\)
b.\(x^2+36=12x\)
\(\Leftrightarrow x^2-12x+36=0\)
\(\Leftrightarrow\left(x-6\right)^2=0\)
\(\Leftrightarrow x-6=0\Leftrightarrow x=6\)
c.\(\frac{1}{16x^2}-x+4=0\)
\(\Leftrightarrow\left(\frac{1}{4x}\right)^2-2.\frac{1}{4x}.2+2^2=0\)
\(\Leftrightarrow\left(\frac{1}{4x}-2\right)^2=0\)
........
Xem thêm câu trả lời
Tìm x biết
a) (x-3)^2 -4=0
b) ( 2x+3)^2 - (2x+1)(2x-1) =22
c) (4x+3)(4x-3) -(4x-5)^2 =16
d) x^3 -9x^2 +27x-27 =-8
e) (x+1)^3 - x^2(x+3) =2
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=0+4\)
\(\left(x-3\right)^2=4\)
\(\left(x-3\right)^2=\pm4\)
\(\left(x-3\right)^2=\pm2^2\)
\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(4x^2+12x+9-4x^2+1=22\)
\(12x+10=22\)
\(12x=22-10\)
\(12x=12\)
\(x=1\)
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
\(16x^2-9-16x^2+40x-25=16\)
\(-34+40x=16\)
\(40x=16+34\)
\(40x=50\)
\(x=\frac{50}{40}=\frac{5}{4}\)
d) \(x^3-9x^2+27x-27=-8\)
\(x^3-9x^2+27x-27+8=0\)
\(x^3-9x^2+27x-19=0\)
\(\left(x^2-8x+19\right)\left(x-1\right)=0\)
Vì \(\left(x^2-8x+19\right)>0\) nên:
\(x-1=0\)
\(x=1\)
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(x^3+2x^2+x+x^2+2x+1-x^2-3x^2=2\)
\(3x+1=2\)
\(3x=2-1\)
\(3x=1\)
\(x=\frac{1}{3}\)
Đúng 0
Bình luận (0)
bài 1:chứng minh rằng với mọi x ta có:
a)-x^2+4x-5<0
b)x^4+3x^2+3>0
c)(x^2+2x+3)(x^2+2x+4)+3>0
bài 2:tìm x:
a)9x^2-6x-3=0
b)x^3+9x^2+27x+19=0
c)x(x-5)(x+5)-(x+2)(x^2-2x+4)=3
Bài 1:
a)-x^2+4x-5
=-(x2-4x+5)<0 với mọi x
=>-x^2+4x-5<0 với mọi x
b)x^4+3x^2+3
\(=\left(x^2+\frac{3}{2}\right)^2+\frac{3}{4}>0\)với mọi x
=>x^4+3x^2+3>0 với mọi x
c) bn xét từng th ra
Bài 2:
a)9x^2-6x-3=0
=>3(3x2-2x-1)=0
=>3x2-2x-1=0
=>3x2+x-3x-1=0
=>x(3x+1)-(3x+1)=0
=>(x-1)(3x+1)=0
b)x^3+9x^2+27x+19=0
=>(x+1)(x2+8x+19) (dùng pp nhẩm nghiệm rồi mò ra)
Với x+1=0 =>x=-1Với x2+8x+19 =>vô nghiệmc)x(x-5)(x+5)-(x+2)(x^2-2x+4)=3
=>x3-25x-x3-8=3
=>-25x-8=3
=>-25x=1
=>x=-11/25
Đúng 0
Bình luận (0)
mk sửa 1 tí ở dấu => thứ 2 từ dưới lên là
=>-25x=11
Đúng 0
Bình luận (0)
tìm x ,biết :
a, x^2-3x = 0
b,x^5-9x = 0
c, ( x^3-4x^2) - ( x-4) = 0
d, (4x^2-25)^2- 9 ( 2x - 5 )^2= 0
a)\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
b)\(x^5-9x=0\)
\(\Leftrightarrow x\left(x^4-9\right)=0\)
\(\Leftrightarrow x\left(x^2+3\right)\left(x^2-3\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x^2+3=0\\x^2-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x\in\varnothing\\x=\pm\sqrt{3}\end{matrix}\right.\)
c)\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x-4\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\x-1=0\\x-4=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\x=1\\x=4\end{matrix}\right.\)
d)\(\left(4x^2-25\right)^2-9\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(2x+5\right)^2\left(2x-5\right)^2-3\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(2x-5\right)^2\left(4x^2+10x+5-3\right)=0\)
\(\Leftrightarrow\left(2x-5\right)^2\left(4x^2+4x+2x+2\right)=0\)
\(\Leftrightarrow\left(2x-5\right)^2\left[4x\left(x+1\right)+2\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(2x-5\right)^2.2\left(2x+1\right)\left(x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-5=0\\2x+1=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}=2,5\\x=-\dfrac{1}{2}=-0,5\\x=-1\end{matrix}\right.\)
Đúng 1
Bình luận (0)
Tìm x biết:
a. 5x(x - 4) - x2 + 16 = 0
b. x + 6x2 + 9x3 = 0
c. x2 - 4x + 3 = 0
a) \(5x\left(x-4\right)-x^2+16=0\)
\(4x^2-20x+16=0\)
\(\Rightarrow\orbr{\begin{cases}x=4\\x=1\end{cases}}\)
Đúng 0
Bình luận (0)
b) \(x+6x^2+9x^2=0\)
\(x\left(3x+1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{3}\end{cases}}\)
Đúng 0
Bình luận (0)
c) \(x^2-4x+3=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Đúng 0
Bình luận (0)