Tìm x :
25-(3x + 5) = 8
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x - (17 - 8) = 5 + (10 - 3x)
x - 17 + 8 = 5 + 10 - 3x
x + 3x = 5 + 10 + 17 - 8
4x = 24
x = 24 : 4
x = 6
b) 25 - (30 + x) = x - (27 - 8)
25 - 30 - x = x - 27 + 8
-x - x = -27 + 8 - 25 + 30
-2x = -14
x = (-14) : (-2)
x = 7
c) (x - 12) - 15 = (20 - 7) - (18 + x)
x - 12 - 15 = 20 - 7 -18 - x
x + x = 20 - 7 - 18 + 12 + 15
2x = 22
x = 22 : 2
x = 11
d) 9 - 25 = (7 - x) - (25 + 7)
9 - 25 = 7 - x - 25 - 7
x = 7 - 25 - 7 - 9 +25
x = -9
Zaza, tự làm nữa đi a~.
a) \(3x\left(x-8\right)+16=2x\)
\(\Rightarrow3x^2-24x+16=2x\)
\(\Rightarrow3x^2-26x+16=0\)
\(\Rightarrow\left(3x^2-24x\right)-\left(2x-16\right)=0\)
\(\Rightarrow3x\left(x-8\right)-2\left(x-8\right)=0\)
\(\Rightarrow\left(x-8\right)\left(3x-2\right)=0\)
Để đẳng thức xảy ra \(\Rightarrow\left[\begin{array}{nghiempt}x-8=0\\3x-2=0\end{array}\right.\)\(\Rightarrow x\in\left\{8;\frac{2}{3}\right\}\)
b) \(\left(5-x\right)^2=25=5^2=\left(-5\right)^2\)
\(\Rightarrow5-x\in\left\{\pm5\right\}\Rightarrow x\in\left\{0;10\right\}\)
1:
a: \(35\cdot16+35\cdot28-44\cdot15\)
\(=35\left(16+28\right)-44\cdot15\)
\(=44\left(35-15\right)\)
\(=44\cdot20=880\)
b: \(240-2\left(3\cdot5^2-20:2^2\right)\)
\(=240-2\left(3\cdot25-20:4\right)\)
\(=240-150+10=10+90=100\)
2:
b: \(\left(8-3x\right)^4-1=15\)
=>\(\left(3x-8\right)^4=16\)
=>\(\left[{}\begin{matrix}3x-8=2\\3x-8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=10\\3x=6\end{matrix}\right.\)
=>x=10/3 hoặc x=2
c: \(218-5\left(x-8\right)=2^5:2^2\)
=>\(218-5\left(x-8\right)=2^3=8\)
=>5(x-8)=210
=>x-8=42
=>x=50
d: \(\left(5-3x\right)^4-1=15\)
=>\(\left(3x-5\right)^4-1=15\)
=>\(\left(3x-5\right)^4=16\)
=>\(\left[{}\begin{matrix}3x-5=-4\\3x-5=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\3x=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\end{matrix}\right.\)
b) ta có \(\left(4x-5\right)^3-\left(2x+5\right)\left(16x^2-25\right)=0\)
\(\left(4x-5\right)^3-\left(2x+5\right)\left(4x+5\right)\left(4x-5\right)=0\)
\(\left(4x-5\right)\left[\left(4x-5\right)^2-\left(2x+5\right)\left(4x+5\right)\right]=0\)
\(\left(4x-5\right)\left(16x^2-40x+5^2-8x^2-10x-20x-5^2\right)=0\)
\(\left(4x-5\right)\left(8x^2-70x\right)=0\)
=> \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}4x=5\\x\left(8x-70\right)=0\end{cases}< =>}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}\) \(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}< =>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}\orbr{\begin{cases}x=\frac{5}{4}\\\orbr{\begin{cases}x=0\\8x-70=0=>x=\frac{35}{4}\end{cases}}\end{cases}}}}\)
\(\orbr{\begin{cases}4x-5=0\\8x^2-70x=0\end{cases}=>\orbr{\begin{cases}x=\frac{5}{4}\\x\left(8x-70\right)=0\end{cases}}}\)
=> \(\orbr{\begin{cases}x=0\\x=\frac{35}{4}\end{cases}}\) Vậy \(\orbr{\begin{cases}x=\frac{5}{4}\\x=0\end{cases}}\) hoặc \(x=\frac{35}{4}\)
a. x - (17 - 8) = 5 + (10 - 3x)
=> x - 9 = 5 + 10 - 3x
=> x + 3x = 15 + 9
=> 4x = 24
=> x = 6
b. 25 - (30 + x) = x - (27 - 8)
=> 25 - 30 - x = x - 19
=> -x -x = -19 + 30 - 25
=> -2x = -14
=> x = 7
c. (x - 12) - 15 = (20 - 7) - (18 + x)
=> x - 12 - 15 = 13 - 18 - x
=> x + x = 13 - 18 + 12 + 15
=> 2x = 22
=> x = 11
a) \(\left(3x-4\right)^3=\left(9x-8\right)\left(3x^2-8\right)\)
\(\Leftrightarrow27x^3+108x^2+144x+64=27x^3-72x-24x^2+64\)
\(\Leftrightarrow27x^3+108x^2+144x+64-27x^3+72x+24x^2-64=0\)
\(\Leftrightarrow132x^2+216x=0\)
\(\Leftrightarrow12x\left(11x+18\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\11x+18=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-\frac{18}{11}\end{array}\right.\)
Bài 1.
\(a, (3x-4)^2\)
\(=\left(3x\right)^2-2\cdot3x\cdot4+4^2\)
\(=9x^2-24x+16\)
\(b,\left(1+4x\right)^2\)
\(=1^2+2\cdot1\cdot4x+\left(4x\right)^2\)
\(=16x^2+8x+1\)
\(c,\left(2x+3\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
\(d,\left(5-2x\right)^3\)
\(=5^3-3\cdot5^2\cdot2x+3\cdot5\cdot\left(2x\right)^2-\left(2x\right)^3\)
\(=125-150x+60x^2-8x^3\)
\(e,49x^2-25\)
\(=\left(7x\right)^2-5^2\)
\(=\left(7x-5\right)\left(7x+5\right)\)
\(f,\dfrac{1}{25}-81y^2\)
\(=\left(\dfrac{1}{5}\right)^2-\left(9y\right)^2\)
\(=\left(\dfrac{1}{5}-9y\right)\left(\dfrac{1}{5}+9y\right)\)
Bài 2.
\(a,\left(x-5\right)^2-\left(x+7\right)\left(x-7\right)=8\)
\(\Rightarrow x^2-2\cdot x\cdot5+5^2-\left(x^2-7^2\right)=8\)
\(\Rightarrow x^2-10x+25-\left(x^2-49\right)=8\)
\(\Rightarrow x^2-10x+25-x^2+49=8\)
\(\Rightarrow\left(x^2-x^2\right)-10x=8-25-49\)
\(\Rightarrow-10x=-66\)
\(\Rightarrow x=\dfrac{33}{5}\)
\(b,\left(2x+5\right)^2-4\left(x+1\right)\left(x-1\right)=10\)
\(\Rightarrow\left(2x\right)^2+2\cdot2x\cdot5+5^2-4\left(x^2-1^2\right)=10\)
\(\Rightarrow4x^2+20x+25-4x^2+4=10\)
\(\Rightarrow\left(4x^2-4x^2\right)+20x=10-25-4\)
\(\Rightarrow20x=-19\)
\(\Rightarrow x=\dfrac{-19}{20}\)
#\(Toru\)
Bài 1
a) (3x - 4)²
= (3x)² - 2.3x.4 + 4²
= 9x² - 24x + 16
b) (1 + 4x)²
= 1² + 2.1.4x + (4x)²
= 1 + 8x + 16x²
c) (2x + 3)³
= (2x)³ + 3.(2x)².3 + 3.2x.3² + 3³
= 8x³ + 36x² + 54x + 27
d) (5 - 2x)³
= 5³ - 3.5².2x + 3.5.(2x)² - (2x)³
= 125 - 150x + 60x² - 8x³
e) 49x² - 25
= (7x)² - 5²
= (7x - 5)(7x + 5)
f) 1/25 - 81y²
= (1/5)² - (9y)²
= (1/5 - 9y)(1/5 + 9y)
25-(3x + 5) = 8
3x+5 = 25-8
3x+5 = 17
3x=17-5
3x= 12
x=12:3
=> x =4
TK MIK NHA
25 - (3x+5) = 8
25 - 3x - 5 = 8
20 - 3x = 8
3x = 12
x = 4