7\(^x\) + 7\(^{x+2}\) = 350
(2x - 3)\(^2\) - 1 = 35
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30 tháng 7 2020
a) 150 : [17 - (2x - 3)] = 350
=> 17 - (2x - 3) = \(\frac{150}{350}=\frac{3}{7}\)
=> 2x - 3 = \(17-\frac{3}{7}=\frac{116}{7}\)
=> 2x = \(\frac{116}{7}+3\)
=> 2x = \(\frac{137}{7}\)
=> x = \(\frac{137}{7}:2=\frac{137}{14}\)
b) 4 . [105 - (x - 9)] - 486 = 0
=> 4. [105 - (x - 9)] = 486
=> 105 - (x - 9) = 243/2
=> x - 9 = \(105-\frac{243}{2}=-\frac{33}{2}\)
=> x = \(-\frac{33}{2}+9=-\frac{15}{2}\)
c) 1000 - 2{600 - 4.[198 - 7.(21 - 7)]}.x = 200
=> 1000 - 2{600 - 4.[198 - 7.14]}.x = 200
=> 1000 - 2{600 - 4.[198 - 98].x = 200
=> 1000 - 2{600 - 4.100}.x = 200
=> 1000 - 2{600 -400}.x = 200
=> 1000 - 2.200.x = 200
=> 2.200.x = 800
=> 400.x = 800
=> x = 2
30 tháng 7 2020
Câu 2 :
a) 350 : [19 - (7x - 3)] = 350
=> 19 - (7x - 3) = 1
=> 7x - 3 = 18
=> 7x = 21
=> x = 3
b) 7.[109 - (x - 9)] - 420 = 0
=> 7.[109 - (x - 9)] = 420
=> 109 - (x - 9) = 60
=> x - 9 = 49
=> x = 49 + 9 = 58
c) 254 - 2{200 - 4.[90 - 7.(19 - 9)]}.x = 54
=> 254 - 2{200 - 4.[90 - 7.10]}.x = 54
=> 254 - 2{200 - 4[90 - 70]} . x= 54
=> 254 - 2{200 - 4.20} . x = 54
=> 254 - 2{200 - 80}.x = 54
=> 254 - 2.120 . x= 54
=> 2.120.x = 200
=> 240.x = 200
=> \(x=\frac{5}{6}\)
27 tháng 6 2021
b) Ta có: \(\left(x^2-7\right)\left(x+2\right)-\left(2x-1\right)\left(x-14\right)+x\left(x^2-2x-22\right)+35\)
\(=x^3+2x^2-7x-14-\left(2x^2-28x-x+14\right)+x^3-2x^2-22x+35\)
\(=2x^3-29x+21-2x^2+29x-14\)
\(=2x^3-2x^2+7\)
4 tháng 7 2016
1) \(\frac{17}{6}-\left(x-\frac{7}{6}\right)=\frac{7}{4}\)
\(\Rightarrow x-\frac{7}{6}=\frac{17}{6}-\frac{7}{4}\)
\(\Rightarrow x=\frac{13}{12}+\frac{7}{6}=\frac{9}{4}\)
2) \(\frac{3}{35}-\left(\frac{3}{5}-x\right)=\frac{2}{7}\)
\(\Rightarrow\)\(\frac{3}{5}-x=\frac{3}{35}-\frac{2}{7}=-\frac{1}{5}\)
\(\Rightarrow x=\frac{3}{5}-\left(-\frac{1}{5}\right)=\frac{4}{5}\)
3) 4) Hjhj^_^^_^
12 tháng 7 2021
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
PT
Phạm Trần Hoàng Anh
CTVHS
25 tháng 6 2021
a) 3/35 - (3/5 + x) = 2/7
=> 3/5 + x= 3/35- 2/7
=> 3/5 +x = -1/5
=> x = -1/5 -3/5
=> x = -4/5
b) 3/7 +1/7 : x = 3/14
=> 1/7 : x= 3/14 -3/7
=> 1/7 : x = -3/14
=> x = 1/7 : -3/14
=> x = -2/3
c) (5x-1).(2x-1/3)=0
=> \(\left[{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}5x=0+1=1\\2x=0+\dfrac{1}{3}=\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{3}:2=\dfrac{1}{6}\end{matrix}\right.\)
Học tốt :D
25 tháng 6 2021
a)x=-4/5
b)x=-2/3
c)\(\left\{{}\begin{matrix}5x-1=0\\2x-\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x=1\\2x=\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\x=\dfrac{1}{6}\end{matrix}\right.\)
Vậy.........
mik lười mong bn thông cảm
25 tháng 6 2023
1:
=>2x-3=0 hoặc 5/2-x=0
=>x=3/2 hoặc x=5/2
2: =>x=1/2+12=12,5
3: =>(2x+3/5-3/5)(2x+3/5+3/5)=0
=>2x(2x+6/5)=0
=>x=0 hoặc x=-3/5
4: =>-1/6x=-1/3
=>x=1/3:1/6=2
5: =>1/4:x=1/4
=>x=1
6: =>2/5x+11/15=1
=>2/5x=4/15
=>x=2/3
16 tháng 6 2023
1: \(=6x^2+2x-15x-5-x^2+6x-9+4x^2+20x+25-27x^3-27x^2-9x-1\)
=-27x^3-18x^2+4x+10
2: =4x^2-1-6x^2-9x+4x+6-x^3+3x^2-3x+1+8x^3+36x^2+54x+27
=7x^3+37x^2+46x+33
5:
\(=25x^2-1-x^3-27-4x^2-16x-16-9x^2+24x-16+\left(2x-5\right)^3\)
\(=8x^3-60x^2+150-125+12x^2-x^3+8x-60\)
=7x^3-48x^2+8x-35
\(7^x+7^{x+2}=350\)
\(\Leftrightarrow7^x\left(1+7^2\right)=350\)
\(\Leftrightarrow7^x.50=350\)
\(\Leftrightarrow7^x=7\)
\(\Leftrightarrow7^x=7^1\)
\(\Leftrightarrow x=1\)
Vậy...
\(\left(2x-3\right)^2-1=35\)
\(\Leftrightarrow\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-3\right)^2=6^2\\\left(2x-3\right)^2=\left(-6\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=6\\2x-3=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
Vậy..
7x +7x+2 =350
↔7x(1+72) = 350
↔7x . 50 = 350
↔ 7x = 7
↔ x =1
( 2x-3 )2 -1 = 35
↔ ( 2x-3)2 = 36 = 62
↔ 2x-3 = 6
↔ 2x=9
↔ x =\(\dfrac{9}{2}\)