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a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Rightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)
\(\Rightarrow-80x=-480\Rightarrow x=6\)
b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)
\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)
\(\Rightarrow15x-8x-5x-6x=36+1-25-12\)
\(\Rightarrow-4x=0\Rightarrow x=0\)
c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Rightarrow10x-16-12x+15=12x-16+11\)
\(\Rightarrow10x-12x-12x=-16+11+16-15\)
\(\Rightarrow-14x=-4\Rightarrow x=\dfrac{2}{7}\)
d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)
\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)
\(\Rightarrow5x-12x+24x-90x+36=182\)
\(\Rightarrow-73x=182-36\)
\(\Rightarrow-73x=146\Rightarrow x=-2\)
Chúc bạn học tốt!!!
a) 4(x+2) - 7(2x - 1) + 9(3x - 4)=30
⇔4x+8 - 14x + 7 + 27x - 36 = 30
⇔ 17x = 51
⇔ x = 3
b) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11
⇔ 10x - 16 - 12x + 15 = 12x - 16 + 11
⇔ -14x = -4
⇔ x= \(\frac{2}{7}\)
c) 5x(1 - 2x) - 3x(x + 18) = 0
⇔ 5x - 10x\(^2\) - 3x\(^2\) -54x =0
⇔ -13x\(^2\) -49 x = 0
⇔ -x ( 13x + 49 ) =0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\13x+49=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-49}{13}\end{matrix}\right.\)
d) 5x - 3{4x - 2[4x - 3(5x - 2)]} = 182
⇔ 5x - 3[ 4x - 2( 4x - 15x + 6 ) ]= 182
⇔5x - 3 ( 4x - 8x + 30x - 12 ) = 182
⇔ 5x - 3 ( 26x - 12 ) = 182
⇔ 5x - 78x + 36 = 182
⇔ - 73x = 146
⇔ x = -2
A. \(4\left(x+2\right)-7\left(2x-1\right)+9\left(3x-4\right)=30\)
\(\Leftrightarrow4x+8-14x+7+27x-36=30\)
\(\Leftrightarrow4x-14x+27x=30-8-7+36\)
\(\Leftrightarrow17x=51\)
\(\Leftrightarrow x=3\) . Vậy \(S=\left\{3\right\}\)
B. \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow10x-12x-12x=16-15-16+11\)
\(\Leftrightarrow10x=-4\)
\(\Leftrightarrow x=-\dfrac{2}{5}\) . Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
Câu C) bạn xem lại đề nha mik tính ko đc
D. \(\left(5x-3\right)4x-2x\left(10x-3\right)=15\)
\(\Leftrightarrow20x^2-12x-20x^2+6x=15\)
\(\Leftrightarrow-6x=15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\) . Vậy \(S=\left\{-\dfrac{5}{2}\right\}\)
a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)
\(\Leftrightarrow72-20x-36x+84=30x-240-6x-84\)
\(\Leftrightarrow156-56x=24x-324\)
\(\Leftrightarrow-80x+480=0\Leftrightarrow x=-6\)
b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x-12\right)+1\)
\(\Leftrightarrow15x+25-8x+12=5x+6x-36+1\)
\(\Leftrightarrow7x+37=11x-35\)
\(\Leftrightarrow-4x+72=0\Leftrightarrow x=18\)
c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\)
\(\Leftrightarrow-14x+4=0\Leftrightarrow x=\frac{2}{7}\)
d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)
\(\Leftrightarrow5x-3\left[4x-15x+6\right]=182\)
\(\Leftrightarrow5x-3\left(-11x+6\right)=182\)
\(\Leftrightarrow5x+33x-18-182=0\)
\(\Leftrightarrow38x-200=0\Leftrightarrow x=\frac{100}{19}\)
\(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-2x-1=12x-5\Leftrightarrow-14x=-4\Leftrightarrow x=\frac{2}{7}\)
\(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-4\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-4x+16\)
\(\Leftrightarrow3x^2-12x-2=3x^2-16x+16\)
\(\Leftrightarrow4x=18\Leftrightarrow x=\frac{9}{2}\)
Bài 2 :
a )
\(\left(4x-3\right)\left(4x+3\right)-15\left(x-1\right)\left(x+1\right)-\left(x+6\right)-3x=1\)
\(\Leftrightarrow16x^2-9-15x^2+15-x-6-3x=1\)
\(\Leftrightarrow x^2-4x-1=0\)
\(\Delta=16+4=20>0\)
\(\Rightarrow\left[{}\begin{matrix}x_1=\dfrac{4+\sqrt{20}}{2}=2+\sqrt{5}\\\dfrac{4-\sqrt{20}}{2}=2-\sqrt{5}\end{matrix}\right.\)
Vậy \(x=2-\sqrt{5}\) hoặc \(x=2+\sqrt{5}\)
b )
\(\left(5x+1\right)\left(5x-1\right)-25\left(x+3\right)\left(x-1\right)=4\)
\(\Leftrightarrow25x^2-1-25x^2-50x+75=4\)
\(\Leftrightarrow-50x+70=0\)
\(\Leftrightarrow x=\dfrac{70}{50}\)
Vậy \(x=\dfrac{70}{50}\)
1) A=x2-4x+4-3=(x-2)2-3
(x-2)2≥0 (Với mọi x)
=> (x-2)2-3 ≥ -3 (V...)
=> Min A=-3
Làm tương tự với những câu khác nha
a: \(A\left(x\right)=2x^4-x^3+3x^2+9x-2\)
\(B\left(x\right)=2x^4-5x^3-x+9\)
\(C\left(x\right)=x^4+4x^2+5\)
A(x): bậc 4; hệ số cao nhất là 2; hệ số tự do là -2
B(x): bậc 4; hệ số cao nhất là 4; hệ số tự do là 9
b: M(x)=A(x)+B(x)=4x^4-6x^3+3x^2+8x+7
N(x)=B(x)-A(x)=-4x^3-3x^2-10x+11
c: Q(x)=-N(x)=4x^3+3x^2+10x-11
Với x = 0 , 0 ko là nghiệm của p/t ( L )
Với x khác 0 , ta có :
\(3x^4-4x^3-5x^2+4x+3=0\)
\(\Leftrightarrow3x^2-4x-5+\frac{4}{x}+\frac{3}{x^2}=0\) ( chia cả 2 vế cho x^2 )
\(\Leftrightarrow3\left(x^2+\frac{1}{x^2}-2\right)-4\left(x-\frac{1}{x}\right)+1=0\)
\(\Leftrightarrow3\left(x-\frac{1}{x}\right)^2-4\left(x-\frac{1}{x}\right)+1=0\)
Đặt \(x-\frac{1}{x}=a\) , ta có :
\(3a^2-4a+1=0\)
\(\Leftrightarrow\left(3a-1\right)\left(a-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{1}{3}\\a=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-\frac{1}{x}=\frac{1}{3}\\x-\frac{1}{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{x^2-1}{x}=\frac{1}{3}\\\frac{x^2-1}{x}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-3=x\\x^2-1=x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3\left(x^2-\frac{x}{3}-1\right)=0\\x^2-x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-\frac{x}{3}-1=0\\\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{1}{6}\right)^2-\frac{37}{36}=0\\\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\frac{1}{6}\right)^2=\frac{37}{36}\\\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pm\sqrt{37}+1}{6}\\x=\frac{\pm\sqrt{5}+1}{2}\end{matrix}\right.\)
Vậy ...
\(3x^4-4x^3-5x^2+4x+3=0\\ \Leftrightarrow3x^4-3x^3-x^3-3x^2+x^2-3x^2+x+3x+3=0\\ \Leftrightarrow3x^2\left(x^2-x-1\right)-x\left(x^2-x-1\right)-3\left(x^2-x-1\right)=0\\ \Leftrightarrow\left(x^2-x-1\right)\left(3x^2-x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2-x-1=0\\3x^2-x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1\pm\sqrt{5}}{2}\\x=\frac{1\pm\sqrt{37}}{2}\end{matrix}\right.\)
`M + 3x^2 - 4 + 5x = x^2 - 4x`
`M = x^2 - 4x - 3x^2 + 4 - 5x`
` M = -2x^2 - 9x + 4`
Vậy ...