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![](https://rs.olm.vn/images/avt/0.png?1311)
a, ĐKXĐ: \(x\ne1;x\ne-1\)
b, Với \(x\ne1;x\ne-1\)
\(B=\left[\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right]\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =\left[\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\right]\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =\dfrac{5}{x^2-1}\cdot\dfrac{4\left(x^2-1\right)}{5}\\ =4\)
=> ĐPCM
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.A=5x-x^2\)
\(=-\left(x^2-5x\right)=-\left[\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\right]=-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}\le\dfrac{25}{4}\)
\(\Rightarrow Max_A=\dfrac{25}{4}\) khi \(x=\dfrac{5}{2}\)
\(b.B=x-x^2=-\left(x^2-x\right)=-\left[\left(x-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\right]=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\)
\(\Rightarrow Max_B=\dfrac{1}{4}\Leftrightarrow x=\dfrac{1}{2}\)
\(c.C=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4-7\right)=-\left(x-2\right)^2+7\le7\)
\(\Rightarrow Max_C=7\Leftrightarrow x=2\)
a) Ta có:
\(A=5x-x^2\)
\(=-\left(x^2-5x\right)\)
\(=-\left(x^2-5x\right)-6,25+6,25\)
\(=-\left(x^2-5x+6,25\right)+6,25\)
\(=-\left(x-2,5\right)^2+6,25\)
Ta lại có:
\(\left(x-2,5\right)^2\ge0\)
\(\Rightarrow-\left(x-2,5\right)^2\le0\)
\(\Rightarrow-\left(x-2,5\right)^2+6,25\le6,25\)
\(\Rightarrow A\le6,25\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2,5\right)^2=0\)
\(\Leftrightarrow x-2,5=0\)
\(\Leftrightarrow x=2,5\)
Vậy MaxA = 6,25 \(\Leftrightarrow x=2,5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=x^2+3x+4=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
\(minA=\dfrac{7}{4}\Leftrightarrow x=-\dfrac{3}{2}\)
b) \(B=2x^2-x+1=2\left(x-\dfrac{1}{4}\right)^2+\dfrac{7}{8}\ge\dfrac{7}{8}\)
\(minB=\dfrac{7}{8}\Leftrightarrow x=\dfrac{1}{4}\)
c) \(C=5x^2+2x-3=5\left(x+\dfrac{1}{5}\right)^2-\dfrac{16}{5}\ge-\dfrac{16}{5}\)
\(minC=-\dfrac{16}{5}\Leftrightarrow x=-\dfrac{1}{5}\)
d) \(D=4x^2+4x-24=\left(2x+1\right)^2-25\ge-25\)
\(minD=-25\Leftrightarrow x=-\dfrac{1}{2}\)
e) \(E=x^2+6x-11=\left(x+3\right)^2-20\ge-20\)
\(minE=-20\Leftrightarrow x=-3\)
f) \(G=\dfrac{1}{4}x^2+x-\dfrac{1}{3}=\left(\dfrac{1}{2}x+1\right)^2-\dfrac{4}{3}\ge-\dfrac{4}{3}\)
\(minG=-\dfrac{4}{3}\Leftrightarrow x=-2\)
\(A=x^2+3x+4=\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{7}{4}=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\)
Do \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\)
\(\Rightarrow A=\left(x+\dfrac{3}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)
\(minA=\dfrac{7}{4}\Leftrightarrow x+\dfrac{3}{2}=0\Leftrightarrow x=-\dfrac{3}{2}\)
Mấy câu còn lại làm tương tự nhé em^^
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1 : a. x^3 +27 -54-x^3 =-27
b. 8x^3 +y^3 -8x^3 +y^3 =2y^3
c. (2x-1+2x+2)(2x-1-2x-2)=(4x+1).(-3)=-12x-3
d. a^3 +b^3 +3ab(a+b) -3ab(a+b)=a^3+b^3
![](https://rs.olm.vn/images/avt/0.png?1311)
c/
Ta có : B=2=>6/2-2x
<=>6=4-4x
<=>6-4=-4x
<=>-4x=2
<=>x=2/-4=-1/2
d/ĐKXĐ:2-2x≠0
<=>2(1-x)≠0<=>-2(x-1)≠0
<=>x≠1
Để giá trị của biểu thức B nguyên thì 2-2x là Ư(6)
=>2-2x ∈ Ư(6)={±1;±2;±3;±6) Nếu 2-2x=1=> -2x=-1=>x=1/2( thoả mãng)
Rồi còn nhiêu bạn tự xét trường hợp y trang cách làm ở trênn nnhan :;)).À sẽ có mấy cái trường hợp nó giống ĐKXĐ thì bạn ghi trong ngoặc ko thoã mãn nhan.
![](https://rs.olm.vn/images/avt/0.png?1311)
a) C được xác định <=> x khác +- 2
b) Ta có : \(C=\dfrac{x^3}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x-1\right)\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=x-1\)
Để C = 0 thì x - 1 = 0 <=> x = 1 (tm)
c) Để C nhận giá trị dương thì x - 1 > 0 <=> x > 1
Kết hợp với ĐK => Với x > 1 và x khác 2 thì C nhận giá trị dương
![](https://rs.olm.vn/images/avt/0.png?1311)
(x+3)2-(4x+1)-2(x+2)
= x2+6x+9-4x-1-2x-4
= x2+(6x-4x-2x)+(9-1-4)
= x2+4
\(A=x^2-4x+1\)
\(A=\left(x^2-4x+4\right)-3\)
\(A=\left(x-2\right)^2-3\)
Vì \(\left(x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x-2\right)^2-3\ge-3\) với mọi x
\(\Rightarrow Amin=-3\Leftrightarrow x=2\)
\(B=4x^2+4x+11\)
\(B=\left(4x^2+4x+1\right)+10\)
\(B=\left(2x+1\right)^2+10\)
Vì \(\left(2x+1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(2x+1\right)^2+10\ge10\) với mọi x
\(\Rightarrow Bmin=10\Leftrightarrow x=-\dfrac{1}{2}\)
\(C=\left(x-1\right)\left(x+3\right)\left(x+2\right)\left(x+6\right)\)
\(C=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(C=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(C=\left(x^2+5x\right)^2-36\)
Vì \(\left(x^2+5x\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x^2+5x\right)^2-36\ge-36\)
\(\Rightarrow Cmin=-36\Leftrightarrow x^2+5x=0\)
\(\Rightarrow x\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(D=5-8x-x^2\)
\(D=-\left(x^2+8x-5\right)\)
\(D=-\left(x^2+8x+16-16-5\right)\)
\(D=-\left(x^2+8x+16\right)+21\)
\(D=-\left(x+4\right)^2+21\)
Vì \(-\left(x+4\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x+4\right)^2+21\le21\) với mọi x
\(\Rightarrow Dmax=21\Leftrightarrow x=-4\)
\(E=4x-x^2+1\)
\(E=-\left(x^2-4x-1\right)\)
\(E=-\left(x^2-4x+4-4-1\right)\)
\(E=-\left(x^2-4x+4\right)+5\)
\(E=-\left(x-2\right)^2+5\)
Vì \(-\left(x-2\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x-2\right)^2+5\le5\) với mọi x
\(\Rightarrow Emax=5\Leftrightarrow x=2\)