3x ( X x 2) = 12x (16- X) +1
Những câu hỏi liên quan
Tìm x biết:
1,
a,3x(x+1) - 2x(x+2) = -x-1
b,2x(x-2020) - x+2020 = 0
c,(x-4)2 - 36 = 0
d,x2 + 8x - 16 = 0
e,x(x+6) - 7x - 42 = 0
f,25x2 - 16 = 0
2,
a,3x3 - 12x = 0
b,x2 + 3x - 10 = 0
Bài 1:
a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)
\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)
b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)
c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
d) \(\Rightarrow\left(x+4\right)^2=0\Rightarrow x=-4\)
e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Bài 2:
a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-2\right)+5\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
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1 Viết các biểu thức sau dưới dạng tích: a) x^2+8x+16 b) x^2-12x+36 c) 4x-4x^2-1 d) x^3-3x^2+3x-1
\(a,=\left(x+4\right)^2\\ b,=\left(x-6\right)^2\\ c,=-\left(4x^2-4x+1\right)=-\left(2x-1\right)^2\\ d,=\left(x-1\right)^3\)
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Tìm x, biết:
a) (2x+2)(x-1)-(x+2)(2x+1)=0;
b)(3x+1)(2x-3)-6x(x+2)=16;
c)(12x-5)(4x-1)+(3x-7)(1-16x)=81
mn ơi giúp mik vs ạ :<
a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0
=>-5x-4=0
=>x=-4/5
b: =>6x^2-9x+2x-3-6x^2-12x=16
=>-19x=19
=>x=-1
c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81
=>83x=83
=>x=1
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26 tháng 10 2021 lúc 9:41
6) \(\sqrt{x^2+12x+36}=-x-6\)
7) \(\sqrt{9x^2-12x+4}=3x-2\)
8) \(\sqrt{16-24x+9x^2}=2x-10\)
9) \(\sqrt{x^2-6x+9}==2x-3\)
10) \(\sqrt{x^2-3x+\dfrac{9}{4}}=\dfrac{3}{x}x-4\)
6) ĐKXĐ: \(x\le-6\)
\(\sqrt{\left(x+6\right)^2}=-x-6\Leftrightarrow\left|x+6\right|=-x-6\)
\(\Leftrightarrow x+6=x+6\left(đúng\forall x\right)\)
Vậy \(x\le-6\)
7) ĐKXĐ: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow\sqrt{\left(3x-2\right)^2}=3x-2\Leftrightarrow\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x-2=3x-2\left(đúng\forall x\right)\)
Vậy \(x\ge\dfrac{2}{3}\)
8) ĐKXĐ: \(x\ge5\)
\(pt\Leftrightarrow\sqrt{\left(4-3x\right)^2}=2x-10\)\(\Leftrightarrow\left|4-3x\right|=2x-10\)
\(\Leftrightarrow4-3x=10-2x\Leftrightarrow x=-6\left(ktm\right)\Leftrightarrow S=\varnothing\)
9) ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow\sqrt{\left(x-3\right)^2}=2x-3\Leftrightarrow\left|x-3\right|=2x-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2x-3\left(x\ge3\right)\\x-3=3-2x\left(\dfrac{3}{2}\le x< 3\right)\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
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28 tháng 6 2023 lúc 10:58
Giải các phương trình sau:
1) \(\sqrt{2x+4}-2\sqrt{2-x}=\dfrac{12x-8}{\sqrt{9x^2+16}}.\)
2) \(\sqrt{3x^2-7x+3}-\sqrt{x^2-2}=\sqrt{3x^2-5x-1}-\sqrt{x^2-3x+4}.\)
tinh gia tri cua bieu thuc:
A=x^2+12x+36 tai x=64
B=x^2+4xy+4y^2 khi biet x=2,8; y=3,6
C=(3x-7)^2+10(3x-7)+25 biet x=16
D=8x^3-12x^2+6x-1 tai x=\(\dfrac{-1}{2}\)
\(A=x^2+12x+36=\left(x+6\right)^2\)
\(B=x^2+4xy+4y^2=\left(x+2y\right)^2\)
\(C=\left(3x-7\right)^2+10\left(3x-7\right)+25=\left(3x-2\right)^2\)
\(D=8x^3-12x^2+6x-1=\left(2x-1\right)^3\)
Việc còn lại bạn tự thay vào rồi tính thôi :v
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\(A=x^2+12x+36\)
\(A=x^2+2.x.6+6^2\)
\(A=\left(x+6\right)^2\)
Thay x = 64 ta được
\(A=\left(64+6\right)^2\)
\(A=70^2\)
\(A=4900\)
\(B=x^2+4xy+4y^2\)
\(B=x^2+2.x.2y+\left(2y\right)^2\)
\(B=\left(x+2y\right)^2\)
Thay x = 2,8 và y = 3,6 ta được
\(B=\left(2,8+2.3,6\right)^2\)
\(B=\left(2,8+7,2\right)^2\)
\(B=10^2\)
\(B=100\)
\(C=\left(3x-7\right)^2+10\left(3x-7\right)+25\)
\(C=\left(3x-7\right)^2+2.\left(3x-7\right).5+5^2\)
\(C=\left(3x-7+5\right)^2\)
\(C=\left(3x-2\right)^2\)
Thay x = 16 ta được
\(C=\left(3.16-2\right)^2\)
\(C=\left(48-2\right)^2\)
\(C=46^2\)
\(C=2116\)
\(D=8x^3-12x^2+6x-1\)
\(D=\left(2x\right)^3-3.\left(2x\right)^2+3.\left(2x\right)-1^3\)
\(D=\left(2x-1\right)^3\)
Thay x = -1/2 ta được
\(D=\left[2.\left(-\dfrac{1}{2}\right)-1\right]^3\)
\(D=\left(-1-1\right)^3\)
\(D=\left(-2\right)^3\)
\(D=-8\)
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11, (6x+5)2(3x+2)(x+1) - 35
12, (4x+1)(12x-1)(3x+2)(x+1)-4
13, (x2+4x+3)(x2+12x+35)+15
14, (x2+13x+40)(x2+15x+54) - 40
15, (x2- 6x - 16)(x2 - 4x - 21) - 144
Tìm x biết
a) x − 1 6 − 6 16 = 25 %
b) 3 x − 1 − 1 2 x + 5 = 0
Bài 3: Tìm x
a) (2x+3)2−4x2=10
b) (x+1)2−(2+x)(x−2)=0
c) (5x−1)(1+5x)=25x2−7x+15
d) (4−x)2−16=0
e) 3x2−12x=0
g) x2−8x−3x+24=0
e: \(\Leftrightarrow3x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
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