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Mấy bài kia phá tung tóe rồi rút gọn hết sức xong thay x vào, làm câu c thôi nhé:
c) \(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
riêng câu này ta thay x = 9 vào luôn, vậy ta có:
\(C=9^{14}-10\cdot9^{13}+10\cdot9^{12}-10\cdot9^{11}+...+10\cdot9^2-10\cdot9+10\)
\(=9^{14}-\left(9+1\right)\cdot9^{13}+\left(9+1\right)\cdot9^{12}-\left(9+1\right)\cdot9^{11}+...+\left(9+1\right)\cdot9^2-\left(9+1\right)\cdot9+10\)
\(=9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-9^{12}-9^{11}+...+9^3+9^2-9^2-9+10\)
\(=-9+10\)
\(=1\)
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
a. \(2x\left(x+5\right)-x\left(3+2x\right)=26\Leftrightarrow2x^2+10x-3x-2x^2=26\Leftrightarrow7x=26\Leftrightarrow x=\dfrac{26}{7}\)
Vậy \(x=\dfrac{26}{7}\)
b. \(5x\left(x-1\right)=x-1\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
c. \(2\left(x+5\right)-x^2-5x=0\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
d. \(\left(2x-3\right)^2-\left(x+5\right)^2=0\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=8\\x=-\dfrac{2}{3}\end{matrix}\right.\)
e. \(3x^3-48x=0\Leftrightarrow3x\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}3x=0\\x^2-16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=\pm4\end{matrix}\right.\)
f. \(x^3+x^2-4x=4\Leftrightarrow x^3+x^2-4x-4=0\Leftrightarrow\left(x^2-4x+4\right)+\left(x^3-8\right)=0\Leftrightarrow\left(x-2\right)^2+\left(x-2\right)\left(x^2+2x+4\right)=0\Leftrightarrow\left(x-2\right)\left(x-2+x^2+2x+4\right)=0\left(x-2\right)\left(x^2+3x+2\right)=0\Leftrightarrow\left(x-2\right)\left(x^2+x+2x+2\right)=0\Leftrightarrow\left(x-2\right)\left[x\left(x+1\right)+2\left(x+1\right)\right]=0\Leftrightarrow\left(x-2\right)\left(x+1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\\x=-2\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-1\\x=\pm2\end{matrix}\right.\)
g. \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
h. \(x^2-4x+8=2x-1\Leftrightarrow x^2-4x+8-2x+1=0\Leftrightarrow x^2-6x+9=0\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy \(x=3\)
__________________________Chúc bạn học tốt____________________________
a) \(\left(y-1\right)^2=9\)
\(\Rightarrow\left(y-1\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
\(\Rightarrow x-1=-3\Rightarrow x=-2\)
Vậy: \(x=4\) hoặc \(-2\)
1/ \(A=3\left(x+1\right)^2-\left(x+3\right)^2\)
\(=3\left(x^2+2x+1\right)-\left(x^2+6x+9\right)\)
\(=3x^2+6x+3-x^2-6x-9\)
\(=2x^2-6\)
Vậy biểu thức A vẫn phụ thuộc vào biến -_-
2/ \(B=\left(x-2\right)^2-\left(x-4\right)x\)
\(=x^2-4x+4-x^2-4x\)
\(=4\)
Vậy biểu thức B không phụ thuộc vào biến (đpcm)
3/ \(C=3\left(x+2\right)^2-3\left(x^2-4x\right)\)
\(=3\left(x^2+4x+4\right)-3x^2+12x\)
\(=3x^2+12x+12-3x^2+12x\)
\(=24x+12\)
Vậy biểu thức C vẫn phụ thuộc vào biến -_-
4/ \(D=3x\left(x-2\right)\left(x+2\right)-x\left(3x+3\right)\)
\(=3x\left(x^2-4\right)-3x^2-3x\)
\(=3x^3-12x-3x^2-3x\)
\(=3x^3-3x^2-15x\)
Vậy biểu thức D vẫn phụ thuộc vào biến -_-
5/ \(E=x^2-\left(x+1\right)\left(x-1\right)+5\)
\(=x^2-\left(x^2-1\right)+5\)
\(=x^2-x^2+1+5\)
\(=6\)
Vậy biểu thức E không phụ thuộc vào biến.
a. 3.(x-2)+2.(x-3)=13
x=5
b. (x+1).(2-x)-(3x+5).(x+2)=-4x2+1
x=-9/10
c.x.(5-2x)+2x.(x-1)=13
x=13/3
d. (2x+3)2-(x-1)2=0
x=-2/3
e. x2.(3x-2)-8+12=0
x vô ngiệm
f x2+x=0
x=-1
g. x3-5x=0
x=0
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
a) \(3\left(x-2\right)+2\left(x-3\right)=1\)\(3\)
\(3x-6+2x-6=13\)
\(5x=13+6+6\)
\(5x=25\)
\(x=25\)
c) \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
d) \(\left(2x+3\right)^2-\left(x-1\right)^2=0\)
\(\left(2x+3-x+1\right)\left(2x+3+x-1\right)=0\)
\(\left(x+4\right)\left(3x+2\right)=0\)
\(\orbr{\begin{cases}x+4=0\\3x+2=0\end{cases}}=>\orbr{\begin{cases}x=-4\\x=\frac{-2}{3}\end{cases}}\)
f) \(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(=>\orbr{\begin{cases}x=0\\x+1=0\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
g) \(x^3-5x=0\)
\(x^2\left(x-5\right)=0\)
\(=>\orbr{\begin{cases}x^2=0\\x-5=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x=5\end{cases}}\) \(\)
\(\)
a) ( 3x - 1 ) ( 2x + 7 ) - ( x + 1 ) ( 6x + 5 ) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 - 5x + 6x - 5) = 16
<=> 6x2 + 21x - 2x - 7 - ( 6x2 + x - 5 ) = 16
<=> 6x2+ 21x - 2x - 7 - 6x2 -x + 5 = 16
<=> 18x - 2 = 16
<=> 18x = 18
=> x = 1
Vậy....
\(a,2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{3;-\dfrac{5}{2}\right\}\)
\(b,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)
\(\Leftrightarrow-\left(3x-2\right)\left(x+11\right)-\left(3x-2\right)\left(2-5x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(-x-11-2+5x\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(4x-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{13}{4}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{\dfrac{2}{3};\dfrac{13}{4}\right\}\)
\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-\dfrac{1}{2};3\right\}\)
\(d,\left(x-1\right)\left(2x-1\right)=x\left(1-x\right)\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1+x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{1;\dfrac{1}{3}\right\}\)
\(e,0,5x\left(x-3\right)=\left(x-3\right)\left(1,5x-1\right)\)
\(\Leftrightarrow0,5x\left(x-3\right)-\left(x-3\right)\left(1,5x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(0,5x-1,5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{1;3\right\}\)
\(f,\left(x+2\right)\left(3-4x\right)=x^2+4x=4\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-x^2-4x-4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-2;\dfrac{1}{5}\right\}\)
\(g,\left(2x^2+1\right)\left(4x-3\right)=\left(x-12\right)\left(2x^2+1\right)\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(x-12\right)\left(2x^2+1\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\forall x\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2+1>0\\x=-3\end{matrix}\right.\)
Vậy nghiệm của pt là \(S=\left\{-3\right\}\)
\(h,2x\left(x-1\right)=x^2-1\)
\(\Leftrightarrow2x\left(x-1\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy nghiệm của pt là \(S=\left\{1\right\}\)
một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?
Mình giải từ cuối lên , mình giải dần -)
n, <=> x(2x-1)-3(2x-1)=0
<=> (x-3)(2x-1)=0
<=> x= 3 hoặc x= 1/2
m, <=> (x+2)(x2-3x+5)-x2(x+2)=0
<=> (x+2)(x2-3x+5-x2)=0
<=> (x+2)(5-3x)=0
=> x= -2 hoặc5/3
\(e,\Leftrightarrow x^2-10x+21-x^2=1\\ \Leftrightarrow-10x=-20\Leftrightarrow x=2\\ f,\Leftrightarrow x^2+4x+4-x^2+3x-2=0\\ \Leftrightarrow7x=-2\Leftrightarrow x=-\dfrac{2}{7}\\ g,=x^2-2x+1-x^2+4=-2x+5\\ h,\Leftrightarrow10x^2+9x-10x^2-13x+3=8\\ \Leftrightarrow-4x=5\Leftrightarrow x=-\dfrac{5}{4}\)
e) \(\left(x-3\right)\left(x-7\right)-x^2=1\\ \Rightarrow x^2-10x+21-x^2=1\\ \Rightarrow-10x=-20\\ \Rightarrow x=2\)
f) \(\left(x+2\right)^2-\left(x-2\right)\left(x-1\right)=0\\ \Rightarrow x^2+4x+4-x^2+3x-2=0\\ \Rightarrow7x=-2\\ \Rightarrow x=-\dfrac{2}{7}\)
g) \(\left(x-1\right)^2-\left(x-2\right)\left(x+2\right)=0\\ \Rightarrow x^2-2x+1-x^2+4=0\\ \Rightarrow-2x=-5\\ \Rightarrow x=\dfrac{5}{2}\)
h) \(\left(10x+9\right).x-\left(5x-1\right)\left(2x+3\right)=8\\ \Rightarrow10x^2+9x-10x^2-13x+3=8\\ \Rightarrow-4x=5\\ \Rightarrow x=-\dfrac{5}{4}\)