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a)
`a-10>b-10`
`<=>a-10+10>b-10+10`
`<=>a>b`
c)
`-a-9≥-b-9`
`<=>-a-9+9≥-b-9+9`
`<=>-a≥-b`
`<=>-a*(-1)/1≤-b*(-1)/1`
`<=>a≤b`
e)
`-4a+9< -4b+9`
`<=>-4a+9-9< -4b+9-9`
`<=>-4a< -4b`
`<=>-4a*(-1)/4> -4b*(-1)/4`
`<=>a>b`
b)
`25+a>25+b`
`<=>25+a-25>25+b-25`
`<=>a>b`
f)
cái giữa là dấu gì vậy ạ
\(a,a-10>b-10\)
\(\Rightarrow a-10+10>b-10+10\)
\(\Leftrightarrow a>b\)
\(b,-a-9\ge-b-9\)
\(\Rightarrow-a-9+9\ge-b-9+9\)
\(\Leftrightarrow-a\ge-b\)
\(c,-4a+9< -4b+9\)
\(\Rightarrow-4a+9-9< -4b+9-9\)
\(\Leftrightarrow a< b\)
\(d,25+a>25+b\)
\(\Rightarrow25+a-25>25+b-25\)
\(\Leftrightarrow a>b\)
Câu cuối thiếu dấu bạn ơi!
Bài 8:
Ta có: \(A=-x^2+2x+4\)
\(=-\left(x^2-2x-4\right)\)
\(=-\left(x^2-2x+1-5\right)\)
\(=-\left(x-1\right)^2+5\le5\forall x\)
Dấu '=' xảy ra khi x=1
\(a,\Leftrightarrow\left(x-9\right)^2-2\left(x-9\right)+1=0\\ \Leftrightarrow\left(x-9-1\right)^2=0\Leftrightarrow x=10\\ b,Sửa:49x^2-14x\sqrt{5}+5=0\\ \Leftrightarrow\left(7x-\sqrt{5}\right)^2=0\Leftrightarrow x=\dfrac{\sqrt{5}}{7}\)
a: \(2x\left(x^2-3x+1\right)=2x^3-6x^2+2x\)
b: \(\left(x+2\right)^2-x^2=4x+4\)
c: \(\left(x+3\right)\left(x^2-3x+9\right)-x^3=27\)
Bài 1.
\(a\Big) 9(4x+3)^2=16(3x-5)^2\\\Leftrightarrow 9[(4x)^2+2\cdot 4x\cdot3+3^2]=16[(3x)^2-2\cdot3x\cdot5+5^2]\\\Leftrightarrow9(16x^2+24x+9)=16(9x^2-30x+25)\\\Leftrightarrow 144x^2+216x+81=144x^2-480x+400\\\Leftrightarrow (144x^2-144x^2)+(216x+480x)=400-81\\\Leftrightarrow 696x=319\\\Leftrightarrow x=\dfrac{11}{24}\\Vậy:x=\dfrac{11}{24}\\---\)
\(b\Big)(x-3)^2=4x^2-20x+25\\\Leftrightarrow(x-3)^2=(2x)^2-2\cdot2x\cdot5+5^2\\\Leftrightarrow(x-3)^2=(2x-5)^2\\\Leftrightarrow (x-3)^2-(2x-5)^2=0\\\Leftrightarrow (x-3-2x+5)(x-3+2x-5)=0\\\Leftrightarrow (-x+2)(3x-8)=0\\\Leftrightarrow \left[\begin{array}{} -x+2=0\\ 3x-8=0 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} -x=-2\\ 3x=8 \end{array} \right.\\\Leftrightarrow \left[\begin{array}{} x=2\\ x=\dfrac{8}{3} \end{array} \right.\\Vậy:...\)
\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)
\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)
\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)
\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)
2.
\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)
ĐKXĐ là :
\(a\ne0;-3;-2\)
Vs a = 1 ta có:
=> P=3
1.
\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)
a) a chia 3 dư 1 nên a = 3k + 1 (k ∈ ℕ)
b chia 3 dư 2 nên b = 3m + 2 (m ∈ ℕ)
⇒ ab = (3k + 1)(3m + 2)
= 9km + 6k + 3m + 2
= 3(3km + 2k + m) + 2
Do 3(3km + 2k + m) ⋮ 3
⇒ 3(3km + 2k + m) + 2 chia 3 dư 2
Vậy ab chia 3 dư 2
b) a chia 5 dư 1 nên a = 5k + 1 (k ∈ ℕ)
b chia 5 dư 2 nên b = 5m + 2 (m ∈ ℕ)
⇒ ab = (5k + 1)(5m + 2)
= 25km + 10k + 5m + 2
= 5(5km + 2k + m) + 2
Do 5(5km + 2k + m) ⋮ 5
⇒ 5(5km + 2k + m) + 2 chia 5 dư 2
Vậy ab chia 5 dư 2
giúp mình với mọi người :<