\(7,\\ a,A=x^2-4x+3+11=\left(x-2\right)^2+10\ge10\\ \text{Dấu }"="\Leftrightarrow x=2\\ b,B=-\left(4x^2-4x+1\right)+6=-\left(2x-1\right)^2+6\le6\\ \text{Dấu }"="\Leftrightarrow x=\dfrac{1}{2}\\ c,x-y=2\Leftrightarrow x=y+2\\ \Leftrightarrow B=y^2-3x^2=y^2-3\left(y+2\right)^2\\ \Leftrightarrow B=y^2-3y^2-12y-12=-4y^2-12y-12\\ \Leftrightarrow B=-\left(4y^2+12y+9\right)-3=-\left(2y+3\right)^2-3\le-3\\ \text{Dấu }"="\Leftrightarrow y=-\dfrac{3}{2}\Leftrightarrow x=\dfrac{1}{2}\)

\(8,\\ \Leftrightarrow x^3-3x^2+5x+a=\left(x-2\right)\cdot a\left(x\right)\)

Thay \(x=2\Leftrightarrow8-12+10+a=0\Leftrightarrow a=-6\)

Bài 7:

a.

$A=(x-1)(x-3)+11=x^2-4x+3+11=x^2-4x+14$

$=(x^2-4x+4)+10=(x-2)^2+10\geq 10$
Vậy gtnn của $A$ là $10$ khi $x=2$

b.

$B=5-4x^2+4x=6-(4x^2-4x+1)=6-(2x-1)^2\leq 6$

Vậy gtln của $B$ là $6$ khi $2x-1=0\Leftrightarrow x=\frac{1}{2}$

c.

$x-y=2\Rightarrow x=y+2$. Khi đó:

$B=y^2-3x^2=y^2-3(y+2)^2=y^2-(3y^2+12y+12)=-2y^2-12y-12$

$=6-2(y^2+6y+9)=6-2(y+3)^2\leq 6$

Vậy $B_{\max}=6$

Bài 8:

Đặt $f(x)=x^3-3x^2+5x+a$

Theo định lý Bê-du, để $f(x)\vdots x-2$ thì $f(2)=0$

$\Leftrightarrow 6+a=0$

$\Leftrightarrow a=-6$

Bài 8 cách khác:

$x^3-3x^2+5x+a=x^2(x-2)-x(x-2)+3(x-2)+(a+6)$

$=(x-2)(x^2-x+3)+(a+6)$

Vậy $x^3-3x^2+5x+a$ chia $x-2$ có dư là $a+6$

Để phép chia là chia hết thì số dư phải bằng $0$

Tức là $a+6=0$

$\Rightarrow a=-6$

Có : A >= 0 + 8 = 8

Dấu "=" xảy ra <=> 1-x=0 <=> x=1

Vậy GTNN của A = 8 <=> x=1

Có : B < = 15 - 0 = 15

Dấu "=" xảy ra <=> x-7=0 <=> x=7

Vậy GTLN của B = 15 <=> x=7

Tk mk nha

a) A=|1-x|+8

=> A-8=|1-x|

=> Để |1-x| có giá trị nhỏ nhất thì A-8=0

=> 1-x =0 => -x=0-1 => -x= -1 => x=1

=> giá trị nhỏ nhất của biểu thức A là:

          |1-1|+8=0+8=8

  Vậy giá trị nhỏ nhất của biểu thức A là 8

\(a,\frac{15}{34}+\frac{7}{21}+\frac{19}{34}-\frac{20}{15}+\frac{3}{7}\)

\(=>\left(\frac{15}{34}+\frac{19}{34}\right)+\left(\frac{7}{21}+\frac{3}{7}\right)-\frac{20}{15}\)

\(=>1+\frac{16}{21}-\frac{20}{15}\)

\(=>\frac{37}{21}-\frac{20}{15}\)

\(=>\frac{3}{7}\)

\(b,12-8\cdot\left(\frac{3}{2}\right)^3\)

\(=>12-8\cdot\frac{27}{8}\)

\(=>12-27\)

\(=>-15\)

\(c,\left(\frac{1}{9}\right)^{2005}\cdot9^{2005}-96^2:24^2\)

\(=>\left(\frac{1^{2005}^{ }}{9^{2005}}\cdot9^{2005}\right)-\left(96^2:24^2\right)\)

\(=>\left(1^{2005}\right)-16\)

\(=>1-16\)

\(=>-15\)

lỗi r bn ơi

Bạn ghi lại đề đi bạn

\(5,M=a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\\ M=\left(a+b\right)\left[\left(a+b\right)^2-3ab\right]\\ M=1\left(1-3ab\right)=1-3ab\ge1-\dfrac{3\left(a+b\right)^2}{4}=1-\dfrac{3}{4}=\dfrac{1}{4}\\ M_{min}=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 5:

\(a+b=1\Rightarrow a=1-b\)

\(M=a^3+b^3=\left(1-b\right)^3+b^3=1-3b+3b^2-b^3+b^3\)

\(=1-3b+3b^2=3\left(b^2-b+\dfrac{1}{4}\right)+\dfrac{1}{4}=3\left(b-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\)

\(minM=\dfrac{1}{4}\Leftrightarrow a=b=\dfrac{1}{2}\)

Câu 7:

\(a^3+b^3+abc\ge ab\left(a+b+c\right)\)

\(\Leftrightarrow a^3+b^3+abc-ab\left(a+b+c\right)\ge0\)

\(\Leftrightarrow a^3+b^3-a^2b-ab^2\ge0\)

\(\Leftrightarrow a^2\left(a-b\right)-b^2\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2-b^2\right)\ge0\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)(đúng do a,b dương)

Dấu "=" xảy ra \(\Leftrightarrow a=b\)

giúp mình vs

5.

Với mọi a;b ta có: \(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow2a^2+2b^2\ge a^2+b^2+2ab\)

\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\Rightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)

\(M=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=a^2+b^2-ab\)

\(M=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\left(a+b\right)^2=\dfrac{3}{2}\left(a^2+b^2\right)-\dfrac{1}{2}\ge\dfrac{3}{2}.\dfrac{1}{2}-\dfrac{1}{2}=\dfrac{1}{4}\)

\(M_{min}=\dfrac{1}{4}\) khi \(a=b=\dfrac{1}{2}\)

6.

Do \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)=2>0\)

Mà \(a^2-ab+b^2>0\Rightarrow a+b>0\)

Mặt khác với mọi a;b ta có:

\(\left(a-b\right)^2\ge0\Rightarrow a^2+b^2\ge2ab\Rightarrow a^2+b^2+2ab\ge4ab\)

\(\Rightarrow\left(a+b\right)^2\ge4ab\Rightarrow ab\le\dfrac{1}{4}\left(a+b\right)^2\) \(\Rightarrow-ab\ge-\dfrac{1}{4}\left(a+b\right)^2\)

Từ đó:

\(2=a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\ge\left(a+b\right)^3-3.\dfrac{1}{4}\left(a+b\right)^2\left(a+b\right)=\dfrac{1}{4}\left(a+b\right)^3\)

\(\Rightarrow\left(a+b\right)^3\le8\Rightarrow a+b\le2\)

\(N_{max}=2\) khi \(a=b=1\)

7.

Ta có:

\(a^3+b^3+abc=\left(a+b\right)\left(a^2+b^2-ab\right)+abc\ge\left(a+b\right)\left(2ab-ab\right)+abc\)

\(\Rightarrow a^3+b^3+abc\ge ab\left(a+b\right)+abc=ab\left(a+b+c\right)\) (đpcm)

Dấu "=" xảy ra khi \(a=b\)

8.

\(\left|a+b\right|>\left|a-b\right|\Leftrightarrow\left(a+b\right)^2>\left(a-b\right)^2\)

\(\Leftrightarrow a^2+2ab+b^2>a^2-2ab+b^2\)

\(\Leftrightarrow4ab>0\Leftrightarrow ab>0\)

\(\Rightarrow a;b\) cùng dấu

`5`

`a, -7/21 +(1+1/3)`

`=-7/21 + ( 3/3 + 1/3)`

`=-7/21+ 4/3`

`=-7/21+ 28/21`

`= 21/21`

`=1`

`b, 2/15 + ( 5/9 + (-6)/9)`

`= 2/15 + (-1/9)`

`= 1/45`

`c, (9-1/5+3/12) +(-3/4)`

`= ( 45/5-1/5 + 3/12)+(-3/4)`

`= ( 44/5 + 3/12)+(-3/4)`

`= 9,05 +(-0,75)`

`=8,3`

`6`

`x+7/8 =13/12`

`=>x= 13/12 -7/8`

`=>x=5/24`

`-------`

`-(-6)/12 -x=9/48`

`=> 6/12 -x=9/48`

`=>x= 6/12-9/48`

`=>x=5/16`

`---------`

`x+4/6 =5/25 -(-7)/15`

`=>x+4/6 =1/5 + 7/15`

`=> x+ 4/6=10/15`

`=>x=10/15 -4/6`

`=>x=0`

`----------`

`x+4/5 = 6/20 -(-7)/3`

`=>x+4/5 = 6/20 +7/3`

`=>x+4/5 = 79/30`

`=>x=79/30 -4/5`

`=>x= 79/30-24/30`

`=>x= 55/30`

`=>x= 11/6`

\(5)\)

\(A=\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)

\(A=\dfrac{-7}{21}+\dfrac{4}{3}\)

\(A=\dfrac{-7}{21}+\dfrac{28}{21}\)

\(A=1\)

\(--------------\)

\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+\dfrac{-6}{9}\right)\)

\(B=\dfrac{2}{15}+\dfrac{-1}{9}\)

\(B=\dfrac{18}{135}+\dfrac{-15}{135}\)

\(B=\dfrac{1}{45}\)

\(------------\)

\(C=9-\dfrac{1}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)

\(C=\dfrac{44}{5}+\dfrac{3}{12}+\dfrac{-3}{4}\)

\(C=\dfrac{528}{60}+\dfrac{15}{60}+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-15}{20}\)

\(C=\dfrac{83}{10}\)

\(6)\)

\(a)\) \(x+\dfrac{7}{8}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}-\dfrac{7}{8}\)

\(x=\dfrac{104}{96}-\dfrac{84}{96}\)

\(x=\dfrac{5}{24}\)

\(b)\) \(\dfrac{-6}{12}-x=\dfrac{9}{48}\)

\(\dfrac{-1}{2}-x=\dfrac{3}{16}\)

\(x=\dfrac{-1}{2}-\dfrac{3}{16}\)

\(x=\dfrac{-8}{16}-\dfrac{3}{16}\)

\(x=\dfrac{-11}{16}\)

\(c)\) \(x+\dfrac{4}{6}=\dfrac{5}{25}-\left(-\dfrac{7}{15}\right)\)

\(x+\dfrac{4}{6}=\dfrac{5}{25}+\dfrac{7}{15}\)

\(x+\dfrac{4}{6}=\dfrac{75}{375}+\dfrac{105}{375}\)

\(x+\dfrac{4}{6}=\dfrac{12}{25}\)

\(x=\dfrac{12}{25}-\dfrac{4}{6}\)

\(x=\dfrac{72}{150}-\dfrac{100}{150}\)

\(x=\dfrac{-14}{75}\)

\(d)\) \(x+\dfrac{4}{5}=\dfrac{6}{20}-\left(-\dfrac{7}{3}\right)\)

\(x+\dfrac{4}{5}=\dfrac{6}{20}+\dfrac{7}{3}\)

\(x+\dfrac{4}{5}=\dfrac{18}{60}+\dfrac{140}{60}\)

\(x+\dfrac{4}{5}=\dfrac{79}{30}\)

\(x=\dfrac{79}{30}-\dfrac{4}{5}\)

\(x=\dfrac{79}{30}-\dfrac{24}{30}\)

\(x=\dfrac{11}{6}\)

Giải:

\(A=\dfrac{-7}{21}+\left(1+\dfrac{1}{3}\right)\)

\(A=\dfrac{-7}{21}+\dfrac{4}{3}\)

\(A=\dfrac{-7}{21}+\dfrac{28}{21}\)

\(A=\dfrac{21}{21}=1\)

\(B=\dfrac{2}{15}+\left(\dfrac{5}{9}+\dfrac{-6}{9}\right)\)

\(B=\dfrac{2}{15}+\dfrac{-1}{9}\)

\(B=\dfrac{6}{45}+\dfrac{-5}{45}\)

\(B=\dfrac{1}{45}\)

\(C=\left(9-\dfrac{1}{5}+\dfrac{3}{12}\right)+\dfrac{-3}{4}\)

\(C=\left(\dfrac{44}{5}+\dfrac{1}{4}\right)+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-3}{4}\)

\(C=\dfrac{181}{20}+\dfrac{-15}{20}\)

\(C=\dfrac{166}{20}\)

Bài 6:

\(a)x+\dfrac{7}{8}=\dfrac{13}{12}\)

⇔\(x=\dfrac{13}{12}-\dfrac{7}{8}=\dfrac{5}{24}\)

\(b)\dfrac{-6}{12}-x=\dfrac{9}{48}\)

⇔\(x=\dfrac{-6}{12}-\dfrac{9}{48}=\dfrac{-11}{16}\)

\(c)x+\dfrac{4}{6}=\dfrac{5}{25}-\dfrac{-7}{15}\)

\(\Leftrightarrow x+\dfrac{2}{3}=\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{2}{3}-\dfrac{2}{3}=0\)

\(d)x+\dfrac{4}{5}=\dfrac{6}{20}-\dfrac{-7}{3}\)

⇔\(x+\dfrac{4}{5}=\dfrac{79}{30}\)

⇔\(x=\dfrac{79}{30}-\dfrac{4}{5}=\dfrac{11}{6}\)