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Bài 1.
a) ( 7x - 3 )2 - 5x( 9x + 2 ) - 4x2 = 18
<=> 49x2 - 42x + 9 - 45x2 - 10x - 4x2 = 18
<=> -52x + 9 = 18
<=> -52x = 9
<=> x = -9/52
b) ( x - 7 )2 - 9( x + 4 )2 = 0
<=> x2 - 14x + 49 - 9( x2 + 8x + 16 ) = 0
<=> x2 - 14x + 49 - 9x2 - 72x - 144 = 0
<=> -8x2 - 86x - 95 = 0
<=> -8x2 - 10x - 76x - 95 = 0
<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0
<=> ( x + 5/4 )( -8x - 76 ) = 0
<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)
c) ( 2x + 1 )2 + ( 4x - 1 )( x + 5 ) = 36
<=> 4x2 + 4x + 1 + 4x2 + 19x - 5 = 36
<=> 8x2 + 23x - 4 - 36 = 0
<=> 8x2 + 23x - 40 = 0
=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))
Bài 2.
a) x2 - 12x + 39 = ( x2 - 12x + 36 ) + 3 = ( x - 6 )2 + 3 ≥ 3 > 0 ∀ x ( đpcm )
b) 17 - 8x + x2 = ( x2 - 8x + 16 ) + 1 = ( x - 4 )2 + 1 ≥ 1 > 0 ∀ x ( đpcm )
c) -x2 + 6x - 11 = -( x2 - 6x + 9 ) - 2 = -( x - 3 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
d) -x2 + 18x - 83 = -( x2 - 18x + 81 ) - 2 = -( x - 9 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
a) \(x^2-x+1\)
\(=\left(x^2-2.x.\dfrac{1}{2}+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\)
b) \(x^2+2x+2\)
\(=\left(x^2+2x+1\right)+1\)
\(=\left(x+1\right)^2+1>0\forall x\)
c) \(-x^2+4x-5\)
\(=-x^2+4x-4-1\)
\(=-\left(x^2-4x+4\right)-1\)
\(=-\left(x-2\right)^2-1< 0\forall x\)
1)
a) \(3x^3y^2-6x^2y^3+9x^2y^2\)
\(=3x^2y^2\left(x-2y+3\right)\)
b) \(5x^2y^3-25x^3y^4+10x^3y^3\)
\(=5x^2y^3\left(1-5xy+2x\right)\)
a, x(x-1)(x+1)(x+2)=24
[x(x+1)]*[(x-1)(x+2)]=24
(x^2+x)*(x^2+x-2)=24
đặt t=x^2+x;ta đc
t*(t-2)=24
t^2-2t=24
t^2-2t+1=25
(t-1)^2=5^2
(t-1)^2-5^2=0
((t-6)(t+4)=0
t=6 hoặc t= -4
với t=6
thì x^2+x=6 <=> (x+1/2)^2 = 25/4 <=> (x+1/2)^2 = (5/2)^2 <=> (x+1/2)^2 - (5/2)^2 =0
đến đây lại áp dụng HĐT thứ 3 giống như khi tìm t lúc nãy là ra
với t= -4 em tự làm
b, 2x(8x-1)^2 (4x-1)=9 <=> (8x-1)^2*(8x^2-2x)=9
<=> (64x^2-16x+1)*(8x^2-2x)=9
đặt t=(8x^2-2x) => 64x^2-16x =8t
ta đc: (8t+1)*t=9 <=> 8t^2+t-9 = 0 <=> (t-1)(8t+9)=0
c, (21/x^2-4x+10)- x^2+4x-6=0 <=> 21/x^2 - x^2 +4 =0
đảt t=x^2 (t#0)
ta đc: 21/t - t + 4 = 0
quy đồng đc: 21-t^2+4t = 0 (với t # 0)
<=> -(t-2)^2 + 25 =0 <=> 5^2 - (t-2)^2 = 0
d, 2x^4-9x^3+14x^2-9x+2=0
vế trái có tổng các hệ số (2-9+14-9+2)=0 nến có 1 nghiêm x=1
nên phân tích đc nhân tử là (x-1)
2x^4-9x^3+14x^2-9x+2=0 <=> (x-1)(2x^3-7x^2+7x-2)=0
<=> x=1 và 2x^3-7x^2+7x-2=0
PT: 2x^3-7x^2+7x-2=0 cũng có tổng các hệ số (2-7+7-2)=0 nên cũng có 1 nghiệm là 1 => vế trái có thể phân tích đc nhân tử (x-1)
2x^3-7x^2+7x-2=0 <=> (x-1)(2x^2-5x+2)=0
<=> x=1 và 2x^2-5x+2=0
2x^2-5x+2=0 <=> x^2 - (5/2)x + 1 =0
<=> (x-5/4)^2 - 9/16 = 0
<=> (x-5/4)^2 - (3/4)^2 = 0
P/s: Thay bằng a,b,c, cho dễ hiểu nha. Tham khảo nhé ♥ ♥ ♥
(x+2)(x+3)-(x-2)(x+5)=0
=> x2+5x+6-x2-3x+10=0
=>2x+16=0
=>2x=-16
=>x=-8
1) \(A=x^2+2x+2=\left(x+1\right)^2+1\ge1>0\left(\forall x\right)\)
2) \(B=x^2+6x+11=\left(x+3\right)^2+2\ge2>0\left(\forall x\right)\)
3) \(C=4x^2+4x-2=\left(2x+1\right)^2-2\ge-2\) chưa chắc nhỏ hơn 0
4) \(D=-x^2-6x-11=-\left(x+3\right)^2-2\le-2< 0\left(\forall x\right)\)
5) \(E=-4x^2+4x-2=-\left(2x-1\right)^2-1\le-1< 0\left(\forall x\right)\)
1. \(A=x^2+2x+2=\left(x+1\right)^2+1\)
Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow\left(x+1\right)^2+1\ge1\)
=> Đpcm
2. \(B=x^2+6x+11=\left(x+3\right)^2+2\)
Vì \(\left(x+3\right)^2\ge0\forall x\)\(\Rightarrow\left(x+3\right)^2+2\ge2\)
=> Đpcm
3. \(C=4x^2+4x-2=-\left(4x^2-4x+2\right)\)
\(=-\left(4\left(x-\frac{1}{2}\right)^2+1\right)\)
Vì \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\Rightarrow4\left(x-\frac{1}{2}\right)^2+1\ge1\)
\(\Rightarrow-\left(4\left(x-\frac{1}{2}\right)^2+1\right)\le1\)
=> Đpcm
4,5 làm tương tự
a. \(\left(2x-1\right)\left(3x+2\right)\left(5-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+2=0\\5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-2}{3}\\x=5\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{1}{2};\dfrac{-2}{3};5\right\}\)
b. \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(x-4\right)+\left(x-5\right)\left(x-4\right)\)
\(\Leftrightarrow3x\left(x-4\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;4\right\}\)
c. \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
\(\Leftrightarrow\left(4x-1\right)^2-4\left(x+3\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left(4x-1\right)\left(4x-1-4x-3\right)=0\)
\(\Leftrightarrow-4\left(4x-1\right)=0\Leftrightarrow4x-1=0\Leftrightarrow x=\dfrac{1}{4}\)
d. \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)
\(\Leftrightarrow27x^2\left(x+3\right)-12x\left(x+3\right)=0\)
\(\Leftrightarrow x\left(27x-12\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\27x-12=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\\x=-3\end{matrix}\right.\)
\(\Rightarrow S=\left\{0;\dfrac{4}{9};-3\right\}\)
e. \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
\(\Leftrightarrow2\left(3x+1\right)^2-\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(6x+1-x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\7x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{3}\\x=\dfrac{-3}{7}\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{-1}{3};\dfrac{-3}{7}\right\}\)
g. \(\left(2x-1\right)^2=49\)
\(\Leftrightarrow2x-1=7\Leftrightarrow x=4\)
\(a,9x^2-6x-3=0\)
\(\Leftrightarrow9x^2-6x+1-4=0\)
\(\Leftrightarrow\left(3x-1\right)^2=4\)
\(\Rightarrow3x-1=\pm2\)
\(\hept{\begin{cases}3x-1=2\Rightarrow x=1\\3x-1=-2\Rightarrow x=\frac{-1}{3}\end{cases}}\)
Vậy \(x=1\) hoặc \(x=\frac{-1}{3}\)
\(b,x^3+9x^2+27x+19=0\)
\(\Leftrightarrow x^3+9x^2+27x+27-8=0\)
\(\Leftrightarrow\left(x+3\right)^3=8\)
\(\Rightarrow x+3=2\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
\(c,x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=3\)
\(\Leftrightarrow x^3-25x-x^3-8=3\)
\(\Leftrightarrow-25x=11\)
\(\Leftrightarrow x=\frac{-11}{25}\)
Vậy \(x=\frac{-11}{25}\)
\(9x^2-6x-3=0\)
<=> \(\left(3x\right)^2-2.3x.1+1-4=0\)
<=> \(\left(3x-1\right)^2-2^2=0\)
<=> \(\left(3x-3\right)\left(3x+1\right)=0\)
<=> \(\hept{\begin{cases}3x-3=0\\3x+1=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=1\\x=\frac{-1}{3}\end{cases}}\)
\(x^3+9x^2+27x+19\) \(=0\)
<=>\(x^3+x^2+8x^2+8x+19x+19=0\)
<=> \(x^2\left(x+1\right)+8x\left(x+1\right)+19\left(x+1\right)=0\)
<=> \(\left(x^2+8x+19\right)\left(x+1\right)=0\)
mà \(x^2+8x+19>0\)
=> \(x+1=0\)
<=> \(x=-1\)
\(x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
<=> \(x\left(x^2-25\right)-\left(x+2\right)\left(x-2\right)^2=3\)
<=> \(x^3-25x-\left(x^2-4\right)\left(x-2\right)=3\)
<=> \(x^3-25x-\left(x^3-2x^2-4x+8\right)=3\)
<=> \(x^3-25x-x^3+2x^2+4x-8=3\)
<=> \(2x^2-21x-8=3\)
<=> \(2x^2-21x-11=0\)
<=> \(2x^2-22x+x-11=0\)
<=> \(2x\left(x-11\right)+\left(x-11\right)=0\)
<=> \(\left(2x+1\right)\left(x-11\right)=0\)
<=> \(\hept{\begin{cases}2x+1=0\\x-11=0\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{-1}{2}\\x=11\end{cases}}\)
Bài 1:
a)-x^2+4x-5
=-(x2-4x+5)<0 với mọi x
=>-x^2+4x-5<0 với mọi x
b)x^4+3x^2+3
\(=\left(x^2+\frac{3}{2}\right)^2+\frac{3}{4}>0\)với mọi x
=>x^4+3x^2+3>0 với mọi x
c) bn xét từng th ra
Bài 2:
a)9x^2-6x-3=0
=>3(3x2-2x-1)=0
=>3x2-2x-1=0
=>3x2+x-3x-1=0
=>x(3x+1)-(3x+1)=0
=>(x-1)(3x+1)=0
b)x^3+9x^2+27x+19=0
=>(x+1)(x2+8x+19) (dùng pp nhẩm nghiệm rồi mò ra)
c)x(x-5)(x+5)-(x+2)(x^2-2x+4)=3
=>x3-25x-x3-8=3
=>-25x-8=3
=>-25x=1
=>x=-11/25
mk sửa 1 tí ở dấu => thứ 2 từ dưới lên là
=>-25x=11