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\(1,A=\left(3x+7\right)\left(2x+3\right)-\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\\ =6x^2+23x+21-2x-3-6x^2-23x+55\\ =73-2x\left(đề.sai\right)\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ 2,\\ a,\Leftrightarrow30x^2+18x+3x-30x^2=7\\ \Leftrightarrow21x=7\Leftrightarrow x=\dfrac{1}{3}\\ b,\Leftrightarrow-63x^2+78x-15+63x^2+x-20=44\\ \Leftrightarrow79x=79\Leftrightarrow x=1\\ c,\Leftrightarrow\left(x+5\right)\left(x^2+3x+2\right)-x^3-8x^2=27\\ \Leftrightarrow x^3+3x^2+2x+5x^2+15x+10-x^3-8x^2=27\\ \Leftrightarrow17x=17\Leftrightarrow x=1\)
\(d,\Leftrightarrow7x-2x^2-3+x^2+x-6=-x^2-x+2\\ \Leftrightarrow9x=11\Leftrightarrow x=\dfrac{11}{9}\)
\(A=x^2-2x+10\)
\(A=\left(x^2-2x+1\right)+9\)
\(A=\left(x-1\right)^2+9\)
Mà \(\left(x-1\right)^2\ge0\)
\(\Rightarrow A\ge9\)
Dấu "=" xảy ra khi :
\(x-1=0\Leftrightarrow x=1\)
Vậy Min A = 9 khi x = 1
\(B=x^2-5x-7\)
\(B=\left(x^2-5x+\frac{25}{4}\right)-\frac{53}{4}\)
\(B=\left(x-\frac{5}{2}\right)^2-\frac{53}{4}\)
Mà \(\left(x-\frac{5}{2}\right)^2\ge0\)
\(\Rightarrow B\ge-\frac{53}{4}\)
Dấu "=" xảy ra khi :
\(x-\frac{5}{2}=0\Leftrightarrow x=\frac{5}{2}\)
Vậy \(B_{Min}=-\frac{53}{4}\Leftrightarrow x=\frac{5}{2}\)
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
a)
\(\left(3x^2-x+1\right)\left(x-1\right)+x^2\left(4-3x\right)=\frac{5}{2}\)
\(\Leftrightarrow3x^3-x^2+x-3x^2+x-1+4x^2-3x^3=\frac{5}{2}\)
\(\Leftrightarrow2x-1=\frac{5}{2}\Leftrightarrow2x=1+\frac{5}{2}=\frac{7}{2}\Leftrightarrow x=\frac{7}{4}\)
b)
\(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)
\(\Leftrightarrow4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8=11\)
\(\Leftrightarrow8x+4-4x+1+8=11\Leftrightarrow4x+13=11\Leftrightarrow4x=-2\Leftrightarrow x=-\frac{1}{2}\)
c)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-7^2\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow-4x+1+6x+9+245=0\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{255}{2}\).
a ) ( 3x2 - x + 1 ) ( x + 1 ) + x2 ( 4 - 3x ) = 5/2
=> 3x3 + 3x2 - x2 - x + x + 1 + 4x2 - 3x3 = 5/2
=> 6x2 + 1 = 5/2
=> 6x2 = 1,5
=> x2 = 0,25
=> x = 0,5
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)
Vậy...
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
3: \(-\dfrac{1}{2}+\left|\dfrac{1}{3}x+\dfrac{2}{5}\right|=\dfrac{5}{7}\cdot\dfrac{3}{11}+\dfrac{8}{11}\cdot\dfrac{5}{7}=\dfrac{5}{7}\)
=>|1/3x+2/5|=5/7+1/2=17/14
=>1/3x+2/5=17/14 hoặc 1/3x+2/5=-17/14
=>x=171/70 hoặc x=-339/70