(5x+3)28 =167
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28(29-23) - 29(28-23)
= 812-644-812+667
= 23
(-167).(67-34)-67.(34-167)
= -11189+5678-2278+11189
= 3400
28(29-23) - 29(28-23)
= 812-644-812+667
= 23
(-167).(67-34)-67.(34-167)
= -11189+5678-2278+11189
= 3400
1 ) 28 + ( 2x - 3 ) = 45
=> 2x - 3 = 17
=> 2x = 20
=> x = 10
2) 167 + 3 ( 15 - x ) =170
=> 3 ( 15 - x ) = 3
=> 15 - x = 1
=> x = 14
3 ) 233 - 9x = 54 : 5
=> 233 - 9x = 53
=> 9x = 233 - 53
=> 9x = 108
=> x =12
28 + [ 2x - 3 ] = 45
2x - 3 = 45 - 28
2x - 3 = 17
2x = 17 + 3
2x = 20
x = 20 : 2
x = 10
167 + 3 . [15 - x ] = 170
3.[15-x] = 170 - 167
3.[15-x] = 33
15 -x = 33 : 3
15 - x = 11
x = 15 -11
x =4
233 - 9 . x = 125
9 . x = 233 - 125
9 . x =108
x = 108 : 9
x = 12
a)(167+35)+(134-167-35) b)(43-31+57)-(243-345+54) c)(89-76)+(76-89+28) d)(243+46-345)-(243-345+54)
=167+35+134-167-35 =43-31+51-43+345-54 =89-76+76-89+28 =243+46-345-243+345-54
=167-167+35-35+134 =12+51-43+345-54 =0+28 =289-345-243+345-54
=0+0+134 =63-43+291 =28 =(-56)-243+345-54
=134 =20+291=311 =(-299)+345-54
=46-54=-8
a ) 1 4 + 3 4 + 2 8 + − 5 13 + 8 13 = 2 8
b ) − 7 31 + 7 31 + 24 19 + − 5 19 + 1 15 = 16 15 c ) − 46 43 + 3 43 + 5 17 + 11 17 + 1 17 = 0 d ) 21 31 + 10 31 + 44 53 + 9 53 + − 16 7 = − 2 7 e ) 15 16 + − 15 16 + 13 33 + 7 33 + 10 33 + 3 33 = 1
a: =>x/27+1=-2/3
=>x/27=-5/3
=>x=-45
b: \(\Leftrightarrow x-4=\dfrac{2}{5}:\dfrac{20}{21}=\dfrac{2}{5}\cdot\dfrac{21}{20}=\dfrac{42}{100}=\dfrac{21}{50}\)
=>x=221/50
c: \(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{4}{60}=\dfrac{1}{15}\)
=>x=1/15-2/3=1/15-10/15=-9/15=-3/5
d: \(\Leftrightarrow x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{15}{14}\cdot\dfrac{21}{20}\)
=>\(x\cdot\dfrac{3}{5}=\dfrac{1}{5}-\dfrac{3}{2}\cdot\dfrac{3}{4}=\dfrac{1}{5}-\dfrac{9}{8}=\dfrac{-37}{40}\)
=>x=-37/24
e: =>-3/7x=84/45
=>x=-196/45
f: =>11/10x=-2/3
=>x=-20/33
\(\sqrt{-3x^3+5x+14}+\sqrt{-5x^3+6x+28}=\left(4-2x-x^2\right)\sqrt{2-x}\) (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 \(S=\left\{2;-1\right\}.\)
(5x + 3)28 = 167
(5x+3)28 = (24)7
(5x+3)28 = 228
5x + 3 = 2
5x = 2 - 3
5x = -1
x = -1/5
(5x+3)28 = (24)7 = -(24)7
\(\Rightarrow\) 5x+3 = 2 = -(2)
\(\left[{}\begin{matrix}5x+3=2\\5x+3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=-1\\5x=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=-1\end{matrix}\right.\)
Vậy x = \(\dfrac{-1}{5}\) hoặc x = -5