Tìm x biết:
|3-8x|<2
Giúp mik vs, chiều nay mik f đi học r! Thanks you.
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\(x\left(x^2+x+1\right)=8\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x-8\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vô-nghiệm\right)\end{matrix}\right.\)
1/8x + 3/8x - 5/3 = 2
=> (1/8 + 3/8)x = 2+5/3
=> 1/2x = 11/3
=> x = 11/3:1/2
=> x = 22/3
Vậy x = 22/3
\(x^4-8x^3+11x^2+8x-12=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-6\right)\left(x-2\right)=0\)
\(\Leftrightarrow x=\left\{1;-1;6;2\right\}\)
\(x^4-8x^3+11x^2+8x-12=0\)
\(\Leftrightarrow x^4-x^3-7x^3+7x^2+4x^2-4x+12x-12=0\)
\(\Leftrightarrow\left(x^4-x^3\right)-\left(7x^3-7x^2\right)+\left(4x^2-4x\right)+\left(12x-12\right)=0\)
\(\Leftrightarrow x^3\left(x-1\right)-7x^2\left(x-1\right)+4x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+4x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2-8x^2-8x+12x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)[x^2\left(x+1\right)-8x\left(x+1\right)+12\left(x+1\right)]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\)x - 1 =0 ; x + 1 = 0 ; x - 2 =0 hoặc x - 6 = 0
\(\Leftrightarrow\)x = 1 ; x = -1 ; x = 2 ; x=6
Ta có 3 x 2 + 8x + 5 = 0
ó 3 x 2 + 3x + 5x + 5 = 0 ó 3x(x + 1) + 5(x + 1) = 0
ó (3x + 5)(x + 1) = 0
Vậy x = - 5 3 ; x = - 1
Đáp án cần chọn là: A
\(8x\left(x-3\right)-8\left(x-1\right)\left(x+1\right)=20\)
\(\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8=20\)
\(\Leftrightarrow\left(8x^2-8x^2\right)+\left(-24x+8\right)=20\)
\(\Leftrightarrow-24x=20-8\)
\(\Leftrightarrow-24x=12\)
\(\Leftrightarrow x=12:\left(-24\right)\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Vậy: \(x=-\frac{1}{2}\)
\(8x\left(x-2\right)-8\left(x-1\right)\left(x+1\right)=20\)
\(\Leftrightarrow8x^2-24x-8\left(x^2-1\right)=20\)
\(\Leftrightarrow8x^2-24x-\left(8x^2-8\right)=20\)
\(\Leftrightarrow8x^2-24x-8x^2+8=20\)
\(\Leftrightarrow-24x+8=20\)
\(\Leftrightarrow-24x=20-8\)
\(\Leftrightarrow-24x=12\)
\(\Leftrightarrow x=-\frac{1}{2}\)
\(2x^3+x^2-8x-4=0\)
\(\Leftrightarrow\)\(x^2\left(2x+1\right)-4\left(2x+1\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(x-2\right)\left(x+2\right)=0\)
đến đây bạn làm tiếp nha
\(2x^3+x^2-8x-4=0\)
\(x^2\left(2x+1\right)-4\left(2x+1\right)=0\)
\(\left(x^2-4\right)\left(2x+1\right)=0\)
\(1.x^2-4=0\)
\(\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow x=\pm2\)
\(2.2x+1=0\)
\(x=-\frac{1}{2}\)
7/5 x X + 3/8 x X= 9
X x (7/5 +3/8)=9
X x 71/40 = 9
X = 9:71/40
X = 360/71
\(\frac{7}{15}X+\frac{3}{8}X=9\)
\(\left(\frac{7}{15}+\frac{3}{8}\right)X=9\)
\(\left(\frac{56}{120}+\frac{45}{120}\right)X=9\)
\(\frac{101}{120}X=9\)
\(X=9:\frac{101}{120}\)
\(X=9x\frac{120}{101}\)
\(X=\frac{1080}{101}\)
Help me, mik hứa ai giúp mik mik sẽ k cho 3k lun đó
=> giá trị tuyệt đối của 3-8x thuộc : 0,1
=0 : 3-8x=0
=>8x=3
=>3/8
=1 : 3-8x=3 3-8x=-3
=>8x=0 =>8x=-6
=>x=0 =>x=-6/8= - 3/4
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