3x-32>-5x+1
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3x+2 / 5x+7 = 3x-1 / 5x + 1
=> (3X+2)( 5X+1) = ( 3x-1) ( 5x+7)
=> 3x ( 5x+1) + 2( 5x+1) = 3x( 5x+7) - 5x - 7
15x^2 + 3x + 10x + 2 = 15x^2 + 21x - 5x - 7
15x^2+ 3x + 10x - 15x^2 - 21x + 5x = -7-2
13x - 16x = -9
-3x = -9
x = 3
Vậy x=3
Đúng 1000%
\(\left(3x-1\right)^5=32\\ \Rightarrow\left(3x-1\right)^5=2^5\\ \Rightarrow3x-1=2\\ \Rightarrow x=1\)
a: \(\Rightarrow10x^2+9x-\left(10x^2+15x-2x-3\right)=8\)
\(\Leftrightarrow10x^2+9x-10x^2-13x+3=8\)
=>-4x=5
hay x=-5/4
b: \(\Leftrightarrow21x-15x^2-35+25x+15x^2-10x+6x-4-2=0\)
=>42x=41
hay x=41/42
a) 4x + 3x = 217
x( 4 + 3 ) = 217
7x = 217
x = 217 : 7 = 31
Vậy x = 31
b) 9x - 3x = 216
( 9 -3)x = 216
6x = 216
x = 216:6 = 36
Vậy x = 36
c) 6x - 3x + 23 = 230
( 6 - 3 )x = 230 - 23
3x = 207
x = 207 : 3 = 69
Vậy x = 69
d) 5x + 3x + x = 72
5x + 3x + 1x = 72
( 5 + 3 + 1 )x = 72
9x = 72
x = 72 : 9 = 8
Vậy x = 8
Chúc bạn học tốt nhé
a) \(4x+3x=217\)
\(\Rightarrow x\cdot\left(3+4\right)=217\)
\(\Rightarrow7x=217\)
\(\Rightarrow x=\dfrac{217}{7}\)
\(\Rightarrow x=31\)
b) \(9x-3x=216\)
\(\Rightarrow x\cdot\left(9-3\right)=216\)
\(\Rightarrow6x=216\)
\(\Rightarrow x=\dfrac{216}{6}\)
\(\Rightarrow x=36\)
c) \(6x-3x+23=230\)
\(\Rightarrow x\cdot\left(6-3\right)=230-23\)
\(\Rightarrow3x=207\)
\(\Rightarrow x=\dfrac{207}{3}\)
\(\Rightarrow x=69\)
d) \(5x+3x+x=72\)
\(\Rightarrow x\cdot\left(5+3+1\right)=72\)
\(\Rightarrow9x=72\)
\(\Rightarrow x=\dfrac{72}{9}\)
\(\Rightarrow x=8\)
\(3x-32>-5x+1\)
\(\Leftrightarrow3x+5x>32+1\)
\(\Leftrightarrow8x>33\)
\(\Leftrightarrow x>\frac{33}{8}\)
\(\Leftrightarrow9x+3=-20x+16\)
=>29x=13
hay x=13/29
\(\sqrt{-3x^3+5x+14}+\sqrt{-5x^3+6x+28}=\left(4-2x-x^2\right)\sqrt{2-x}\) (ĐKXĐ: \(x\in R,x\le2\))
\(\Leftrightarrow\sqrt{\left(2-x\right)\left(3x^2+6x+7\right)}+\sqrt{\left(2-x\right)\left(5x^2+10x+14\right)}-\left(4-2x-x^2\right)\sqrt{2-x}=0\)
\(\Leftrightarrow\sqrt{2-x}\left(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}-4+2x+x^2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\left(1\right)\end{cases}}\)
Pt \(\left(1\right)\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-\left(x+1\right)^2+5\left(2\right)\)
Ta có: \(\left(x+1\right)^2\ge0\Rightarrow\sqrt{2\left(x+1\right)^2+4}\ge\sqrt{4}=2\)
Tương tự: \(\sqrt{5\left(x+1\right)^2+9}\ge3\). Từ đó: \(VT_{\left(2\right)}\)\(\ge2+3=5\)
Mà \(VP_{\left(2\right)}=-\left(x+1\right)^2+5\le5\) nên dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)(tm)
Vậy tập nghiệm của pt cho là \(S=\left\{2;-1\right\}.\)
tui bấm lộn vào ta6
\(3x-32>-5x+1\)
\(\Leftrightarrow3x+5x>1+32\)
\(\Leftrightarrow8x>33\)
\(\Leftrightarrow x>\frac{33}{8}\)