Help me!
Tìm x:
3x + 3x+3 = 756
5x+1 + 6 . 5x+1 = 875
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TN
3
28 tháng 6 2017
Câu 1:
\(M=x^2-3x+5\)
\(M=x^2-2.\frac{3}{2}x+\frac{9}{4}+\frac{11}{4}\)
\(M=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
Dấu = xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Min M = 11/4 khi x=3/2
b)\(N=2x^2+3x\)
\(N=2\left(x^2+\frac{3}{2}x\right)\)
\(N=2\left(x^2+2.\frac{3}{4}x+\frac{9}{16}\right)-\frac{9}{8}\)
\(N=2\left(x+\frac{3}{4}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)
Dấu = xảy ra khi \(x+\frac{3}{4}=0\Rightarrow x=-\frac{3}{4}\)
Vậy MIn N = -9/8 khi x=-3/4
c)Tự làm nha
28 tháng 6 2017
Ta có : x2 - 3x + 5
= x2 - 2.x.\(\frac{3}{2}\) + \(\frac{3}{2}^2\) + \(\frac{11}{4}\)
= \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\in R\)
Nên : \(\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\) \(\ge\frac{11}{4}\forall x\in R\)
Vậy GTNN của biểu thức là : \(\frac{11}{4}\) khi \(x=\frac{3}{2}\)
3 tháng 2 2017
a. \(3-4x\left(25-2x\right)-8x^2+x-300=0\)
\(\Leftrightarrow3-100x+8x^2-8x^2+x-300=0\)
\(\Leftrightarrow-297-99x=0\)
\(\Leftrightarrow x=3\)
Vậy \(n_0\) của PT là: x=3
b. \(\Leftrightarrow\frac{\left(2-6x\right)}{5}-2+\frac{3x}{10}=7-\frac{3x+3}{4}\)
\(\Leftrightarrow\frac{\left(4-12x\right)}{5}-\frac{20}{10}+\frac{3x}{10}=\frac{\left(28-3x-3\right)}{4}\)
\(\Leftrightarrow\frac{\left(-16-9x\right)}{10}=\frac{\left(25-3x\right)}{4}\)
\(\Leftrightarrow-64-36x=250-30x\)
\(\Leftrightarrow-6x=314\)
\(\Leftrightarrow x=-\frac{157}{3}\)
Vậy -\(n_0\) của PT là: \(x=\frac{-157}{3}\)
c. \(5x+\frac{2}{6}-8x-\frac{1}{3}=4x+\frac{2}{5}-5\)
\(\Leftrightarrow-3x=4x-\frac{23}{5}\)
\(\Leftrightarrow7x=\frac{23}{5}\)
\(\Leftrightarrow x=\frac{23}{35}\)
Vậy \(n_0\) của PT là: \(x=\frac{23}{35}\)
d. \(3x+\frac{2}{3}-3x+\frac{1}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow\frac{5}{6}=2x+\frac{5}{3}\)
\(\Leftrightarrow x=-\frac{5}{12}\)
Vậy \(n_0\) của Pt là: \(x=-\frac{5}{12}\)
LN
24 tháng 7 2017
1) x3 + 5x2 + 3x - 9
= x3 + 2x2 + 3x2 + 6x - 3x - 9
= ( x3 + 2x2 ) + (3x2 + 6x ) - ( 3x + 9 )
= x2 ( x+ 2 ) + 3x ( x + 2) - 3( x +2 )
= ( x + 2 ) ( x2 + 3x -3 )
2) x3 + 5x2 + 8x + 4
= ( x3 + x2 ) + ( 4x2 + 4x ) + ( 4x + 4 )
= x2 ( x + 1 ) + 4x ( x + 1 ) + 4 ( x + 1 )
= ( x + 1) ( x2 + 4x + 4 )
= (x + 1 ) ( x + 2 )2
3) x3 - 9x2 + 6x + 16
= x3 - 8x2 - x2 + 8x - 2x + 16
= ( x3 - 8x2 ) - ( x2 - 8x ) - ( 2x - 16 )
= x2 ( x - 8 ) - x ( x - 8 ) - 2 ( x - 8 )
= ( x - 8 ) ( x2 - x - 2 )
4) x3 - 4x2 + x + 6
= x3 - 3x2 - x2 + 3x - 2x + 6
= ( x3 - 3x2 ) - ( x2 - 3x ) - ( 2x - 6)
= x2 ( x - 3 ) - x ( x- 3 ) - 2 ( x - 3)
= ( x - 3 ) ( x2 - x - 2 )
NM
2
25 tháng 11 2023
ĐKXĐ: \(x\in R\)
\(3x^2-5x+6=2x\cdot\sqrt{x^2-x+2}\)
=>\(3x^2-6x+x-2+8=2\cdot\sqrt{x^4-x^3+2x^2}\)
=>\(\left(x-2\right)\left(3x+1\right)=2\cdot\left(\sqrt{x^4-x^3+2x^2}-4\right)\)
\(\Leftrightarrow\left(x-2\right)\left(3x+1\right)=2\cdot\dfrac{x^4-x^3+2x^2-16}{\sqrt{x^4-x^3+2x^2}+4}\)
=>\(\left(x-2\right)\left(3x+1\right)=2\cdot\dfrac{x^4-2x^3+x^3-2x^2+4x^2-8x+8x-16}{\sqrt{x^4-x^3+2x^2}+4}\)
=>\(\left(x-2\right)\left(3x+1\right)=\dfrac{2\left(x-2\right)\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\)
=>\(\left(x-2\right)\left[\left(3x+1\right)-\dfrac{2\left(x^3+x^2+4x+8\right)}{\sqrt{x^4-x^3+2x^2}+4}\right]=0\)
=>x-2=0
=>x=2(nhận)
25 tháng 11 2023
\(3x^2-5x+6=2x\sqrt{x^2-x+2}\)
\(\Leftrightarrow\left[x^2-2x\sqrt{x^2-x+2}+\left(x^2-x+2\right)\right]+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{x^2-x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{x^2-x+2}\\x-2=0\end{matrix}\right.\Leftrightarrow x=2\)
Thử lại ta thấy nghiệm \(x=2\) thỏa phương trình ban đầu.
17 tháng 7 2016
\(\Rightarrow\left(3x+2\right).\left(5x-3\right)=\left(5x+7\right)\left(3x-1\right)\)
\(\Rightarrow15x^2-9x+10x-6=15x^2-5x+21x-7\)
\(\Rightarrow19x-6=26x-7\)
\(\Rightarrow26x-19x=7-6\)
\(\Rightarrow13x=1\)
\(\Rightarrow x=\frac{1}{13}\)
PH
2
T
25 tháng 7 2018
\(a̸\)
\(\frac{3}{5}-\frac{1}{2}.x=\frac{1}{4}\)
\(\frac{1}{2}.x=\frac{3}{5}-\frac{1}{4}\)
\(\frac{1}{2}.x=\frac{7}{20}\)
\(\Rightarrow\frac{7}{10}\)
\(b̸\)
\(11,3+2\left[x-\frac{1}{3}\right]=\frac{25}{6}\)
\(2\left[x-\frac{1}{3}\right]=\frac{25}{6}-11,3\)
\(2\left[x-\frac{1}{3}\right]=\frac{-107}{15}\)
\(x-\frac{1}{3}=\frac{-107}{15}:2\)
\(x-\frac{1}{3}=\frac{-107}{30}\)
\(x=\frac{-107}{30}+\frac{1}{3}\)
\(x=\frac{-97}{30}\)
25 tháng 7 2018
\(a)\frac{3}{5}-\frac{1}{2}x=\frac{1}{4}\)
\(\implies\frac{1}{2}x=\frac{3}{5}-\frac{1}{4}\)
\(\implies\frac{1}{2}x=\frac{7}{20}\)
\(\implies x=\frac{7}{20}:\frac{1}{2}\)
\(\implies x=\frac{7}{10}\)
Vậy...
\(b) 11,3+2(x-\frac{1}{3})=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x-2.\frac{1}{3}=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x-\frac{2}{3}=\frac{25}{6}\)
\(\implies \frac{113}{10}+2x=\frac{25}{6}+\frac{2}{3}\)
\(\implies \frac{113}{10}+2x=\frac{29}{6}\)
\(\implies 2x=\frac{29}{6}-\frac{113}{10}\)
\(\implies 2x=\frac{-97}{15}\)
\(\implies x=\frac{-97}{30}\)
Vậy..
\(c)5x-435+2x+140+3x=565\)
\(\implies (5x+2x+3x)+(-435+140)=565\)
\(\implies 10x+(-295)=565\)
\(\implies 10x=565-(-295)\)
\(\implies 10x=860\)
\(\implies x=86\)
Vậy...
~ hok tốt a~
3x + 3x+3 = 756
=> 3x + 3x.33 = 756
=> 3x + 3x.27 = 756
=> 3x.(1 + 27) = 756
=> 3x.28 = 756
=> 3x = 756 : 28
=> 3x = 27 = 33
=> x = 3
5x+1 + 6.5x+1 = 875
=> 5x+1.(1 + 6) = 875
=> 5x+1.7 = 875
=> 5x+1 = 875 : 7
=> 5x+1 = 125 = 53
=> x + 1 = 3
=> x = 3 - 1
=> x = 2
a)
=> 3x(1+33)=756
=>3x=27=33
=>x=3
câu b tương tự