a)X - 1/4 - 5/3 x1/2 b)X:2/3-1/2+3/4
sắp hết hạn làm bài rồi giúp mik nhé
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a)x : 3 - 7 * 2 = 1,56
x : 3 - 14 = 1,56
x : 3 = 1,56 + 14
x : 3 = 15,56
x = 15,56 * 3
x = 46,68
b)nếu dấu chấm trên đề là dấu phẩy thì:
x : 6,4 = 1,248
x = 1,248 * 6,4
x = 7,9872
c)(x + 1) + (x + 2) + (x + 3) + (x + 4) + (x + 5) = 40
x + 1 + x + 2 + x + 3 + x + 4 + x + 5 = 40
(x + x + x + x + x) + (1 + 2 + 3 + 4 + 5) = 40
5 * x + 15 = 40
5 * x = 40 - 15
5 * x = 25
x = 25 : 5
x = 5
A) X:3 - 7 * 2 = 1,56
X:3-14=1,56
X:3=1,56+14
X:3=5,56
X= 5,56* 3 =16,68
B) X : 6 . 4 = 1 ,248
X = 1,248 :4 *6
X=1,872
C) ĐỀ DÀI LÀM LUN
= 5X + (1+2+3+4+5) =40
5X+15 =40
5X= 40-15=25
X= 25:5=5
a) 2(x + 3) = 5(1 - x) - 2
<=> 2x + 6 = 5 - 5x - 2
<=> 2x + 6 = 3 - 5x
=> 2x - 5x = 6 + 3
=> -3x = 9
=> x = 9 : (-3)
=> x = -3
7: Ta có: \(\left(3x+4\right)\left(2x-1\right)+6x\left(1-x\right)=0\)
\(\Leftrightarrow6x^2-3x+8x-4+6x-6x^2=0\)
\(\Leftrightarrow11x=4\)
hay \(x=\dfrac{4}{11}\)
8: Ta có: \(2x\left(x^2-1\right)+x\left(-2x^2-3x+1\right)=-x-27\)
\(\Leftrightarrow2x^3-2x-2x^3-3x^2+x+x+27=0\)
\(\Leftrightarrow x^2=9\)
hay \(x\in\left\{3;-3\right\}\)
x2-4x+4=4x2-12x+9
\(\Leftrightarrow\)3x2-8x+5=0
\(\Leftrightarrow\)3x2-3x-5x+5=0
\(\Leftrightarrow\)3x(x-1)-5(x-1)=0
\(\Leftrightarrow\)(x-1)(3x-5)=0
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5}{3}\\x=1\end{cases}}\)
b,x2-2x-25=0
\(\Leftrightarrow\)(x-1)2-26=0
\(\Leftrightarrow\)(x-1-\(\sqrt{26}\))(x-1+\(\sqrt{26}\))=0
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{26}+1\\x=-\sqrt{26}+1\end{cases}}\)
2, a, x^2-2x+1+4=(x-1)^2+4\(\ge\)4
b, 4x^2-4x+1-1+y^2+2y+1-1-2015=(2x-1)^2+(y+1)^2-2017\(\ge\)-2017
mk làm như thế thôi chứ bài kia dài quá mk làm biếng sory
Nguyễn Thị Hà Tiên : Cảm ơn bạn nhiều lắm =)) Mik đã bt hướng làm bài rồi :3 Thực sự cảm ơn pạn nek <3
Bài 1:
a) \(\left(x-2\right)^2=4x^2-12x+9\Leftrightarrow\left(x-2\right)^2=\left(2x-9\right)^2\Leftrightarrow\left(x-2\right)^2-\left(2x-9\right)^2=0\)
\(\Leftrightarrow\left(x-2+2x-9\right)\left(x-2-2x+9\right)=0\Leftrightarrow\left(3x-11\right)\left(7-x\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}3x-11=0\Leftrightarrow3x=11\Leftrightarrow x=\frac{11}{3}\\7-x=0\Leftrightarrow-x=-7\Leftrightarrow x=7\end{cases}}\)
VẬy tập nghiệm của phương trình là : S={11/3 ; 7}
b) Nếu x^2 -2x =25 thì lẻ lắm . Tớ nghĩ phải là : x^2 -2x = 24
Bài 2 :
a) \(A=x^2-2x+5=x^2-2x+1+4=\left(x-1\right)^2+4\)
vì \(\left(x-1\right)^2\ge0\) nên \(\left(x-1\right)^2+4\ge4\) hay \(A\ge4\)
Vậy GTNN của A là 4 khi x = 1 ( hay x-1 =0 )
b) \(B=4x^2-4x+y^2+2y-2015=\left(4x^2-4x+1\right)+\left(y^2+2y+1\right)-2017\)
\(=\left(2x-1\right)^2+\left(y+1\right)^2-2017\)
Vì \(\left(2x-1\right)^2\ge0\) và \(\left(y+1\right)^2\ge0\) nên \(\left(2x-1\right)^2+\left(y+1\right)^2-2017\ge-2017\)
HAy \(B\ge-2017\) Vậy GTNN của B là -2017 khi x=1/2 và y = -1
a)\(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)
\(=\left(x^2-9\right)\left(x+2\right)-\left(x^3-3x-x^2+3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)
\(=x^3+2x^2-9x-18-x^3+x^2+3x-3-5x^3-40x^2-80x-x^2+10x-25\)
\(=-5x^3-38x^2-76x-46\)
b)\(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)
\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2\left(x^2+10x+25\right)-\left(x^2-2x+1\right)\)
\(=2x^3-16x^2+32x-\left(x^3+5x^2-4x-20\right)+2x^2+20x+50-x^2+2x-1\)
\(=x^3-20x^2+58x+69\)
c)\(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x+3\right)\left(x-3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3-48x^2-36x-\left(2x^3-x^2-18x+9\right)\)
\(=-18x^3-46x^2-8x+16\).
a) Ta có: \(\left(x-3\right)\left(x+3\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x+4\right)^2-\left(x-5\right)^2\)
\(=\left(x^2-9\right)\left(x+2\right)-\left(x-1\right)\left(x^2-3\right)-5x\left(x^2+8x+16\right)-\left(x^2-10x+25\right)\)
\(=x^3+2x^2-9x-18-\left(x^3-3x-x^2+3\right)-5x^3-40x^2-80x-x^2+10x-25\)
\(=-4x^3-39x^2-79x-43-x^3+3x+x^2-3\)
\(=-5x^3-38x^2-76x-46\)
b) Ta có: \(2x\left(x-4\right)^2-\left(x+5\right)\left(x-2\right)\left(x+2\right)+2\left(x+5\right)^2-\left(x-1\right)^2\)
\(=2x\left(x^2-8x+16\right)-\left(x+5\right)\left(x^2-4\right)+2x^2+20x+50-x^2+2x-1\)
\(=2x^3-16x^2+32x-x^3+4x-5x^2+20+x^2+22x+49\)
\(=x^3-20x^2+56x+49\)
c) Ta có: \(\left(x+5\right)^2-4x\left(2x+3\right)^2-\left(2x-1\right)\left(x-3\right)\left(x+3\right)\)
\(=x^2+10x+25-4x\left(4x^2+12x+9\right)-\left(2x-1\right)\left(x^2-9\right)\)
\(=x^2+10x+25-16x^3+48x-36x-2x^3+18x+x^2-9\)
\(=-18x^3+2x^2+40x+16\)
Bài 1 :
a, \(\frac{3}{4}:x=\frac{5}{12}\)
\(x=\frac{3}{4}:\frac{5}{12}\)
\(x=\frac{9}{5}\)
b, \(x-\frac{1}{2}=\frac{3}{4}:\frac{3}{2}\)
\(x-\frac{1}{2}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{1}{2}\)
\(x=1\)
c, \(1\frac{1}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x-\frac{1}{2}=\frac{3}{4}\)
\(\frac{3}{2}x=\frac{3}{4}+\frac{1}{2}\)
\(\frac{3}{2}x=\frac{5}{4}\)
\(x=\frac{5}{4}:\frac{3}{2}\)
\(x=\frac{5}{6}\)
Bài 2 :
\(A=\frac{-3}{5}+\left(\frac{-2}{5}-99\right)\)
\(A=\frac{-3}{5}+\frac{-2}{5}-99\)
\(A=\left(-1\right)-99\)
\(A=-100\)
\(B=\left(7\frac{2}{3}+2\frac{3}{5}\right)-6\frac{2}{3}\)
\(B=\left(\frac{23}{3}+\frac{13}{5}\right)-\frac{20}{3}\)
\(B=\frac{23}{3}+\frac{13}{5}-\frac{20}{3}\)
\(B=\left(\frac{23}{3}-\frac{20}{3}\right)+\frac{13}{5}\)
\(B=1+\frac{13}{5}\)
\(B=\frac{18}{5}\)
a) \(x.\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)=x.\left(x^2-16\right)-\left(x^4-1\right)=x^3-16x-x^4+1\)
ý này ko rút gọn được hết đâu.
b) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)
\(=y^4-81-y^4+4=-77\)
c) \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc=a^2+b^2+c^2+2ab-2bc-2ac-a^2+2ac-c^2-2ab+2bc=b^2\)
a) \(x-\frac{1}{4}-\frac{5}{3}\text{x}\frac{1}{2}=x-\frac{1}{4}-\frac{5\text{x}1}{3\text{x}2}\)
\(\Rightarrow x-\frac{1}{4}-\frac{5}{6}=x-\left(\frac{1}{4}+\frac{5}{6}\right)\)
\(\Rightarrow x-\left(\frac{3}{12}+\frac{10}{12}\right)=x-\frac{13}{12}\)
b) \(x:\frac{2}{3}-\frac{1}{2}+\frac{3}{4}=\frac{3}{2}x-\frac{1}{2}+\frac{3}{4}\)
\(\Rightarrow\frac{3x}{2}-\frac{1}{2}+\frac{3}{4}=\frac{3x-1}{2}+\frac{3}{4}\)
\(\Rightarrow\frac{\left(3x-1\right)\text{x}2}{4}+\frac{3}{4}=\frac{\left(3x-1\right)\text{x}2+3}{4}\)
\(\Rightarrow\frac{6x-2+3}{4}=\frac{6x+1}{4}\)
mik đng cần siên gấp luôn nên giúp mik